mindstrech - stories about numbers, maths puzzles and games (gnv64).pdf

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Why is the correct abbreviation of Karan Johar's Kabhi Khushi Kabhi Gham 3KG and not K3G?

Which is the smallest multi-digit perfect square, that is also a palindrome, and all of whose digits are odd?

Can you find the product of any two two-digit numbers without multiplying, adding or subtracting?

lis Scared, baffled, panicked... Is that how you reel at the very mention 01 the word 'maths'? Does it make your palms sweat and give you worry lines on your forehead? Well, here's a book that will prove to you that playing with numbers can actually be fun.

Mindstretch invites you to put your skills of problem-solving, deduction, logic and lateral-thinking to the test without being attacked by lemni.i . construction, induction and sundry other mathematkal demons (Villained within diese pages is an incredible collection of brainteaser.s, whodunnits, mystery and logic games that will challenge your mental powei and j>n u you with hours of enthralling entertainment. And iiiteispcisi d w nh < us facts about numbers, are magic squares, riddles on chess, ,i spn ia| cr< >••.« < ml and a fascinating section showing you how to draw your own ni.i/i-

So stretch your mind and experience mathemagic!

Debkumar Mitra

m 1 n

A PENGUIN ORIGINAL Reference

www.penguinbooksindia.com

S t o r i e s a b o u t n u m b e r s , m a t h s p u z z l e s a n d g a m e s

http://www.penguinbooksindia.com

PENGUIN BOOKS MINDSTRETCH

llrlikumar Mitra is a Sunday painter and a failed mathematician. Starting his career as a researcher in mathematical biology, Debkumar noon discovered that he had too many other things on his mind. He joined the Statesman and spent his time there mostly re-reading political n ports and writing science features. He was then with the Telegraph as I'.u t of the launch team for the daily's science supplement Knowhow. A former University of Toronto Gordon M. Fisher fellow, Debkumar i nlisi overed his love for mathematics while writing the puzzles column 'Ki.unstorming' for the supplement. Now research director at Derek < Vllricn &C Associates, he still writes the column, paints, and lives with Ins wife and two lovely children in Kolkata.

Penguin Books India (P) Ltd., 11 Community Centre, Panchsheel Park, New Delhi 110017, India Penguin Books Ltd., 80 Strand, London WC2R ORL, UK Penguin Group Inc., 375 Hudson Street, New York, NY 10014, USA Penguin Books Australia Ltd., 250 Camberwell Road, Camberwell, Victoria 3124, Australia Penguin Books Canada Ltd., 10 Alcorn Avenue, Suite 300, Toronto, Ontar io M4V 3B2, Canada Penguin Books (NZ) Ltd., Cnr Rosedale & Airborne Roads, Albany, Auckland, New Zealand Penguin Books (South Africa) (Pty) Ltd., 24 Sturdee Avenue, Rosebank 2196, South Africa

First published by Penguin Books India 2004 Copyright O Debkumar Mitra 2004

All rights reserved 10 9 8 7 6 5 4 3 2 1

Typeset in Times by Eleven Arts, New Delhi

Printed at Pauls Press, New Delhi

This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out , or otherwise circulated without the publisher's prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights undi i copyright reserved above, no part of this publication may be reproduced, stored ill or introduced into a retrieval system, or transmitted in any form or by any meanv (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the above-mentioned publisher of this book.

To the 'S's in my life— shefali, Shila, Sharmistha, Sree and Sourjya

C o n t e n t s

A U T H O R ' S N O T E i x

R E A L / U N R E A L 1

C E L L U L O I D M A T H 1 1

C A B S , C H A I N S A N D B I R T H D A Y S

G A M E S G A L O R E 2 2

J U D Y ' S L O G I C I 3 3

J U D Y ' S L O G I C II 6 7

N U M B E R T A L E S 1 0 6

P U Z Z L E M A N I A I 1 1 8

P U Z Z L E M A N I A II 1 5 7

X - N U M A N D L A B Y R I N T H S 1 9 8

Author's Note

I IN- nli M of a book like this was born on the day I was rejected by nil interview board in 1986. Throughout the half-an-hour-|lin)i interrogation, the respected mathematicians across the InMf vv i iv trying to gauge how little I knew. As I left that dimly III, |ai(.•on-filled room I was sure what I would never be . . . a lANthcinulics teacher. So, how was I to give back to the subject llinl h.ul given me so much—brilliant insights, the ability to llilnk mi my feet and a profusion of ideas? From that thought, Mlndstrctch was born.

I always loved puzzles. As a young reader I was introduced !•• M.ii tin Gardner's columns in Scientific American. He is still I N S I ' I I I I I I lis columns implored me to look beyond the four walls til ins mathematics classroom and explore the subject beyond A | > O M M | ' S Calculus. He still inspires me to make audacious

lures and prove them too. 111 ii ii- Mindstretch is a supersonic ride through mathematics,

•Oldlng the jargon bushes. It invites readers who care two hoots Ini numbers to join the ride.

Now, it is thank you time. I thank all my teachers for liiiiuihu ing me to the wonderful world of patterns—thank you Miilns Roy, Priyotosh Khan and Amiya Bhushan Roy. It all Mni ii il with my first mathematics teacher, my father-—Baba, this U Ini vim 1 fondly remember Pathik Guha, my editor at the Irli Khi/ili. who encouraged me to start a mathematics column. THntik you Judy, without your fantastic logic puzzles this book S M H I I I I never have seen the day; Sudipta, I simply loved the | * I I / / Irs ,md all of them are here; all readers of the 'Brainstorming'

X A u t h o r ' s N o t e

column—thanks for all the support; and Prasun, thanks for taking-all the pain to edit 'Brainstorming' all these years. And Udayan Mitra of Penguin, thanks for suffering me all the while.

Despite Sakshi Narang's meticulous editing, an odd erroi may still crop up. Bear with it because a mathematician always 'thinks A, writes B and talks about C'—and I am after all a failed mathematician.

Kolkata January 2004

Debkumar Mitra

R e a l / U n r e a l

M< ntpbooks are always full of surprises. Unfinished projects, lilen embryos, silly doodles, bright sparks, and more, they are iill< Mill-prints of a human invention factory. One rarely visits pyrnpbooks unless one is lost in the dark alley of idealessness. I lii' iu-xl lew sections are idea doodles from a scrapbook. These HI. ir.il stories set in unreal places. They are like numbers—so imiriil, vet they help weave the fabric of the touch and feel world nl iinliirc.

Did Pierre de Fermat Have a Proof of the Greatest Pufilo He Ever Constructed?

I'rtiI i>1 ,111 Unfinished Play: Prelude

ft M'ri v exactly ten years ago in 1993; a British mathematician l^ift Inns; nl Princeton University, Andrew Wiles, shook the world

iiimouncing to his peers that he had a proof of the notorious ftn'tiun f ImsI Theorem (FLT).

In n conference held in June 1993 in Cambridge University, tWn k< i vca st •ties of three lectures named elliptic curves, modular

JimM mul Galois representations. At the end of these lectures, M came to a proof of Taniyama-Shimura conjecture for semi-ttilbh clliptK curves. From this proof follows the proof of FLT.

2 M i n d s t r e t c h

For about 350 years before Wiles' announcement, FL'l teased and tormented hordes of brilliant minds. And every time a new attempt had been made, mathematics benefited with new tools being discovered, but FLT could still not be cracked.

There were many who believed Pierre de Fermat's famous conjecture could never be proved or disproved. Despite Fermat's claim that he had a 'marvellous proof', many believed, or still believe, that the 'prince of amateurs' never had a proof. Given the reputation of Fermat, no one, however, disbelieved him.

Fermat lived in Toulouse, southern France, in the yean 1601-65, and was the most productive mathematician of that generation. Fermat is known best as the Father of Number Theory, the branch of mathematics dealing mainly with whole numbers. Fermat was also one of the founders of Analytical Geometry, Calculus and Probability.

Did Fermat have a proof? Here Wiles meets the man himself in his dream to find out. . .

PIERRE-: DE F E R M A T : Cubum autem in duos cubos, aut quadrato-quadratum in duos quadrato-quadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.

(A rough translation from J.T. Bell's Men of Mathematics: On the contrary, it is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree: I have discovered a truly marvellous demonstration [of this general theorem] which this margin is too narrow to contain.)

A N D R E W W I L E S : Sorry, I do not understand Latin. But this sounds familiar...

R e a l / U n r e a l 3

11 MM w It should be. You nurtured this dream... you took the wiml out of my sail.

Wn i . Honestly, I do not understand...

I I MM A I I ook, Englishman, I 'm Pierre.

Win s (looks surprised) I'm so sorry, I could not recognize yon Mi Fermat, I owe this to you.

PMMA'i : You owe nothing to me. You closed a chapter on a Inn)' and agonizing search. We all salute you. (The French Itlw yn extends his hand.)

II lisii'i \ that Fermat mentions: The equation x2 + y2 = z2 has Miinv solutions where x, y, and z are integers, i.e. 32 + 42 = 52, Of 9* * 12- = 132. Such a solution is called a Pythagorean ll ti'li l, because according to the Pythagoras Theorem, such a llll'lit i< presents the sides of a right-angled triangle.

I lie problem of finding Pythagorean triplets is an example hi i »«• < if the types of problems that occupied Diophantus, after whom <uv named the 'Diophantine equations'. In these pifiiiiln >ns, the number of unknowns is greater than the number ill aquation s, and the required solution must be in integers only. Diopluintus was a Greek mathematician of the third century, whi > lived in Alexandria, Egypt. Only half of the thirteen )Hilnni< \ of his book Arithmetica have survived to the present (juv Many books of other Greek mathematicians have suffered 11 similar fate; others subsisted only through their translation to \ ruble.

Ihpatiu of Alexandria published, in the beginning of the Hflh i eiitury, a first commentary to Diophantus' book, but it Mil* lost, like the rest of her writings. Two more commentaries n mi i icuted in the thirteenth century by Byzantines: the Afolo^iiin Georgius Pachymeres, who also wrote a book about

4 2 M i n d s t r e t c h

the history of the Byzantine Empire, and the monk Maximus Planudes. Byzantine scholars, who fled from Constantinople in the middle of the fifteenth century, brought the book to western Europe. It wasn 'tfor another 170 years that the book became known to the public, when in 1621, Claude Bachet published a Latin translation of it. Fermat used a copy of this book.)

W I L E S : Thank you, but you started it all by jotting that famous note on the margins of Claude Bachet's translation of Diophantus.

FERMAT: It must have been a long, lonely search for you. Did you feel the pain of being almost there?

WILES: Let me repeat what I told the BBC in an interview. Perhaps 1 can best describe my experience of doing mathematics in terms of a journey through a dark, unexplored mansion. You enter the first room of the mansion and it's completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is. Finally, after six months or so, you find the light switch, you turn it on, and suddenly it's all illuminated. You can see exactly where you were.

FERMAT: H O W does it feel to solve my little teaser?

W I L E S : After seven years of loneliness, suddenly it became clear to me. And once with the help of my student Richard Taylor I removed a fatal flaw in my initial argument, the proof of Fermat's Last Theorem was there. Thanks again for that little scribble. Are you unhappy that a proof has been found?

FERMAT: Unhappy? Me? On the contrary, I 'm happy that I have been vindicated.

W I L E S : Did you actually prove the theorem?

FERMAT: (gives an uncomfortable smile) What do you think?

W I L E S : Honestly, I have my doubts... I used lots of tools that


wrn developed after your time, for example the Taniyama-Miimuiii conjecture.

11 MM vi Your proof to my conjecture, published in two articles inili. M.iy 1995 issue of the journal Annals of Mathematics, is Well beyond even a professional mathematician and I was only mi itnuiteur. A lawyer with a passing interest...

SN n i Please do not embarrass me.. . but you are sidetracking lite INNUC.

h'MMAi: (looks annoyed) No, I 'm not! I'll go by what I wrote > yeurs ago.

Wtti '. So, you had a proof?

I p i" i i I 'lease read my notes on the margins. (Leaves abruptly.)

II ti urn I ermat, pursued by Wiles.)

Profrssor Moriarty: A Lament or a Flourishing Mathematical Proof Nipped in the Bud

A I llll'1 Note from Biographical Sketches of Mathematicians

J M I I I I ' S Moriarty was a mathematical genius who had great iHfliiem e on many aspects of the Victorian society. He is best fjhown as the criminal adversary of Sherlock Holmes. His |)|ntiiuplin al details are sketchy and the best account of his early lilt i < lli \ career has been an extraordinary one. He is a man Mnoi'd birth and excellent education, and is endowed by Nature Willi 11 phenomenal mathematical faculty. At the age of twenty-Mflf hi w rote a treatise on the binomial theorem, which has had H WffOpean vogue. On the strength of it, he won the mathematical Mlllr ill one of our smaller universities, and had, to all Itpparanci w, a most brilliant career before him. But the man hail In n ilitary tendencies of the most diabolical kind. A


criminal strain ran in his blood, which, instead of being modified, was increased and rendered infinitely more dangerou s by his extraordinary mental powers. Dark rumours gathered around him in the university town, and eventually he was compelled to resign from his chair and to come down to London, where he set up as an army coach. He is known to have had an interest in the applications of Pure Mathematics also: He is the celebrated author of The Dynamics of an Asteroid, a book which ascends to such rarefied heights of pure mathematic s that it is said that there was no man in the scientific press capable of criticizing it.

Why Did Sir Arthur Conan Doyle Portray a Brilliant Mathematician as a Criminal: A Conjecture

Proposition I

Sir Arthur was a medical professional and could not appreciate mathematics so he decided to ignore James Moriarty.

Argument in favour

That Conan Doyle did not know much about mathematics becomes apparent in Dr Watson's description of the legendary mathematician—'At the age of twenty-one, he wrote a treatise on the binomial theorem, which has had a European vogue.' Everything about the binomial theorem can be explained in at most six pages, so no one can write a treatise on this subject. Furthermore, Sir Isaac Newton explained and proved all we need to know about the theorem a hundred years before Moriarty.

Argument against

If Conan Doyle did not appreciate mathematics, then why did he create a mathematics professor? Also, the deductive logic


IImmI l>v Ins pipe-smoking hero to solve many of his cases is hi. Is used by mathematicians.

I'n position II

Nil Aithui was well aware of the power of mathematics and Mlrsr i l the best logical mind conjures up the most difficult of • i inn* plots.

Aifiiinii-nl in favour

I liiiui'li Moriarty does appear in a very few cases, but the way III Wit i son describes 'the Napoleon of crime' is proof enough lliiii < on.in Doyle had a healthy respect for his mathematical mlml A i liminal strain ran in his blood, which, instead of being humlilic<I. was increased and rendered infinitely more dangerous hy Ins <-*iinordinary mental powers.'

AfgUttirnt .igainst (a little weak though)

Til* Vol v absence of Moriarty from a major part of Holmes' in ' indicates that Conan Doyle did not have healthy respect fill Ills criminal's mind. And the deathly embrace at the MIt hrnhiu li Falls and Holmes' triumphant return go against I It* proposition.

Ptopmltlon III

I I I A I I I I I I I vsas confused vis-a-vis the status of mathematics, Mill J I H I U ' S Moriarty suffered in the process.

Insulin it-nl biographical material to build a case for or against llw» pni|Hisilion: further research advised. •• If tine goes through all the propositions and arguments, not

linn Ii I HI he concluded. However, this gives interesting, not ihough, insights into how a mathematical proof can


be constructed. Keep questioning the propositions to get a well rounded view; mathematics is all about discovering patterns and proofs are a way to achieving it.

On Reading Uncle Petros and Goldbach's Conjecture

A Small Dose of Number Folklore

For many amateurs, the dream is to achieve the impossible and one of the toughest hurdles to cross is the notorious Goldbach's Conjecture. Christian Goldbach was a Prussian amateur who chanced upon this discovery. In his 7 June 1742 letter to Leonur11 Euler, he asked to check his assertion. Euler thought it to be loo trivial, but for the past 300 years, Goldbach's demon continues to torment even the best of mathematicians. What did Goldbach ask Euler to check?

'Every even number greater than 4 can be written as the sum of two prime numbers." Is this assertion correct?

Years of Dreaming

One got hooked on to Goldbach at an early age. With no knowlet Ige of how and what of number theory, it was a naive walk through the known. Years of checking whether Goldbach was right, and often fooling one with misplaced notions such as 'since all prime numbers except 2 are odd, the sum of any two such numbers will be an even number'. Obviously, as days changed the colour ol one's hair, the conjecture revealed itself with its entire monstrosil v

Hey, Mr Goldbach, Your Letter is Worth a Million!

Last night was rather long. Finished reading Uncle Petros ami Goldbach's Conjecture by Apostolos Doxiadis. This was 12 June 2002. I reached for the pen and paper: a note in the book said


' " 'IdluK li s Conjecture and take a million dollars'. Cool, I in I Ins was a dream moment for a failed mathematician. Mtlt llii euplioi i.i Listed only an hour; the competition was open itlll v li 11 > 1.1, nl, of the US and the UK. All solvers had to submit • N m w ID ii recognized (read 'refereed') mathematics journal 'Mtu r Illicit'on 15 March 2002. What a shame! I was more tlwn i' "iiplr of months late. Who was I fooling? A proof that glHtU I iil< i lite Crown Prince of Mathematics, will dawn on • IIIMI I'ooh!

iMjA V.i .md After I* ill. Iiesi book on a puzzle that is close to my heart. Uncle

^ H f Will (ioldbach 's Conjecture tells the story of the narrator's M l * , H JtUh'iK Meek mathematician named Petros Papachristos, ABI pilled he. strengths against Goldbach's Conjecture and lost. &HiHil» i ol ii\il life mathematicians appear as characters in D | , MIL h .IS Constantin Caratheodory, G.H. Hardy, J.E. Hill** I Srinivasa Ramanujan, and Kurt Godel.

A pi I'.on.il 11 si of the world's greatest mathematical puzzles A ^ l l t i ' i Marilyn vos Savant, one whose IQ graph went through

| (liildlni. li s Conjecture Hi. /I i hi Primes Conjecture that there are infinitely many

Mm* ol pnmes only 2 units apart I* 'li. i. Iw .iys a prime between n2 and (n + l)2?

M M M * there exist a number that is perfect and odd? 1 I ' i . I 111Uteri's Tenth Problem

l <HH<II<<<I OH the Bottom Margin

H M H tin we p i from here? Evariste Galois' last duel over a M R M I I IIIHI lie. I .II nous letter that opened a whole area of research


in mathematics; or, Poisson folklore: This French mathematician got his famous 'distribution' formula by studying the number of soldiers dying of mule-kicks in the Prussian army; of, schizophrenia and Kurt Godel; or, discovery of Louis Bacheliei s 1900 dissertation, 'The Theory of Speculation', and the death ol Black Scholes formula for minimizing risk on stock market floors; or, logician Gottlob Frege's mental state after receiving Lord Bertrand Russell's letter on 'Russell's Paradox'; or, imagining a world without mathematics.

These stories, anecdotes and speculative tales along with lemmas, theorems and numbers form the body of mathematics Mathematics is a human endeavour, and without a glimpse into the quirky ways of its practitioners, one cannot complete the great tale of numbers and lines.

Celluloid Math

f p i L I M I T H I in of mathematics extends well beyond our cut-«ml ili\ > la .srooms. In two most successful entertainment Mmiit'N "I (In world—Holly wood and Bollywood—mathematical

mi' used and abused in equal measure.

M K. oi I KG

yuii seen the film Kabhi Khushi Kabhi Gham? If you |HVf Hi ii. \ i HI did not miss much. The message from the review S p k m n l I hi' world's one of the most respected cinema magazines, §j§/lI A Snunri, seems to indicate that.

I In l ili n with its paint-the-town-red actor list did not enthuse H i , However, its title in the abbreviated form caught one's iii> muni K HI! HfcprWingly, the title of the film has only one G, in Gham.

J f e (line urc three K words—Kabhi twice and Khushi once. on ciuth then do we get K3G? Is this a new notation for

Dlillllon ' \\ hen we add three Ks, what do we get? 3K or K3? Vltli (In not have to be a great mathematician to know it is 3K; ft* K i K i K (1 + 1 + 1) K = 3K.

No |f11111j* by the non-Bollywood, standard mathematical Miiliillon tin abbreviated title of the film should be 3KG.


Another thought crossed the mind when one looked at this new title—'three kilogram' cannot be a title of a Karan Johar film The total media weight of the stars of the title is way above three kilograms! No wonder, Mr Johar had to reinvent mathematical notation.

By the way, Johar's earlier film Kuch Kuch Hota Hai is popularly known as K2H2. Looks suspiciously like a wrongly written chemical formula for some potassium compound! How does that smell?

A word of caution for the young readers of this book: Please stick to the standard notation while doing algebra because Mr Johar is yet to rewrite our mathematics books.

Now, let us leave Mr Johar with his heavyweights and look at another flavour of the Bollywood kind, Aamir Khan's Lagaan It is puzzle-time in Champaner, the backdrop of the Oscar-nominated film.

So What If It Did Not Win an Oscar

After all the award hooplas were over, the eleven cricket stars of the Lagaan team sat down to count the number of awards. A recent news report claimed that the total number of awards the film won was a four-digit number.

The team found that the number mentioned in the report w as way above what they had got. True to their honest credentials, the team members wrote to the editor: 'Sir, thank you for heaping, those awards on us. However, the total number of awards the film got is a three-digit number and not four as mentioned in your report. Interestingly, the number of awards is divisible by 11 without a remainder. Now, the number that you get by dividing the number of awards by 11 is equal to the sum of the square ol the digits of the number of awards.'

C e l l u l o i d M a t h 13

| he ediloi shot back: 'Nice riddle, Aamir, but tell me which ilu i should I take?' VVlis w as the editor confused?

Number Dealers on Celluloid

If ymii miiul is like a mathematical bird nest, nothing even mtioii Is linked to the subject will escape you. There are many * llit Invi io decorate their 'nests' with mathematical gems, if MH (link < >neol them, Arnold G.Reinhold, collects information Hi fvpi i'senlation of mathematicians or mathematics in motion pi | Icre is a must-see selection from his digital library:

J , (forlrsiux (1974); a TV biography on Rene Descartes by Hoherto Kossellini

) l{> 11 ' I 'M) ; Walter Matthau plays Einstein in this film. Lou i lin obi is ihe Hungarian genius Kurt Godel

* /V h, i Number (1991); the story of a wandering mathematician, Hun |k Paul Erdos, obsessed with unsolved problems. A liuin h ol I amous Hungarian mathematicians also appear in Kiln 111m

4, Infinity (1996); Matthew Broderick's biopic about Richard I ovntnan A HfiniUful Mind (2001); a biopic of game theory specialist, liilm Nash

ft h im,n 'v hist Tango (2001); It is a musical fantasia based nil Mil true encounter between mathematician Andrew Wiles mill format's last theorem. The characters in this film include I II. lid, ( .II I I riedrich Gauss, Pythagoras, Pierre de Fermat .md >ii Isaac Newton Ihtiold • list is endless and fascinating with a lot of

B '


Dear Arnie's Reversible Logic

I always misspell Arnold Schwarzenegger's name. This brain bender is only a pretext to memorize the spelling a few times. Here it goes:

No one knows why this out-take from Schwarzenegger's Terminator II did not even make it to the director's cut. Here is a brief description of the scene. Schwarzenegger enters the Hall of Number Images in search of a code. The room is full of four-digit numbers in the form of mirror images. He is looking for a four-digit number code. He knows he can ignore all such numbers that have their first digit as 0, and his code is a four-digit perfect square such that its mirror image also is a perfect square. Schwarzenegger gets the code in 46 nanoseconds. Can you? By the way, a mirror image of a number is defined as the number obtained by reversing the digits of the original number, for example, if the number is ABCD then its mirror image is DCBA. (Arun K. Dey)

Beyond Reinhold: Some Films

1. Conceiving Ada by Lynn Hershman Leeson (1998): It is an exper imental , independent film about a modern-day mathematician/computer programmer who, through MIT, gets 'in touch' with Ada Byron King, the first computer programmer. With digital effects that are extraordinary for the time, Leeson addresses issues of life and death, self-will and community good with a tight crew and stage.

2. Drowning by Numbers by Peter Greenaway (UK/Netherlands, 1988): Greenaway loves the arcane parlour game. He fashions a plot in which games are the modus operandi. Obsessive game playing provokes the characters, poses intellectual puzzles, and provides an unpredictable order to things.


I, t truth i'/ <i Neapolitan Mathematician by Mario Martone lllllly, ()n the evening of May Day 1959, a drunkard is nlii|i|n'd hv (lie police. Presumed to be a vagrant, he is in fact MwmliK a. 11oppolI i, an esteemed mathematics professor at Naples UlllVtMnllY. pundson of the anarchist Mikhail Bakunin, and a Minimum i I he lilm depicts the last week of the professor's life.

flvr l"i Hell—A Puzzling Look

lit lli# I WW original, Klaus Kinsky stars in this World War II Dim I lit ,II my chooses five of their best to spearhead a suicide mllrnliiii mined al recovering the offensive strategy plans of the i n in II. u- is a different version:

MI MIIH is of 'Talent Extreme' enter an enemy village and ((llit11 he 11 | ii >sitions code-named position 1, position 2, position | Million I and position 5. The team members have different

^ m M i tlllVeivnt colour of hair (or bald); each of them is an expert III II in l.l il operation; and each of them have a professional lllllt'i' All we know about them is:

I /.i | ' | i is in the church.

| 'I In li.dd headed member of the team has positioned lilm i II next to the computer whiz.

H ttli*hy is an expert in picking locks. I I In table is on the left of the pawnbroker shop. Ii'l'lti" one al the stable is an expert grenade thrower.

m 'I he 11 virologist is also a make-up expert. I 'Hi. ventriloquist's neighbouring position is taken up by the

I i i it In line expert. H Miinly is a map-maker,

l i t Ni'fiov is at position one. |U, 'lln Miiinloquist has positioned himselfnext to the member

w h o has salt and pepper hair.


11. Jetto has red hair. 12. The mathematician has curly black hair. 13. The person at position three is a laser beam expert. 14. Neeroy's position is next to the hospital. 15. The one at the school is a computer whiz.

Talent Extreme won the battle of wits hands down. Tell us what is the profession of the blonde? (This is a re-creation of a famous puzzle.)

Mathematics Curio in Films

• The star figure used to symbolize Ashtalakshmi, the eight forms of wealth, appears prominently in the Lugash National Museum portrayed in the film Return of the Pink Panther

Figure 1

Snake Lemma:

A - B D

a

O • A ' * B1 - C 1

Figure 2


A tltiifi.mi lemma which states that the above commutative tllntliiun ol Abelian groups and group homeomorphisms with •NVl lows gives rise to an exact sequence.

In (he o|>ening scene of Claudia Weill's film It's My Turn 11 UN! 11 ,i loniautic comedy, Jill Clayburgh, playing a mathematics |i|nl. HIII 'I proves the 'snake lemma' to an obnoxious graduate • l I l U l The rest of the film is mostly free of mathematics and lull "i Mi> had Douglas.

• I MM I.II continuous walk:

Figure 3

HlltM I I I I I I I Ik Avengers, Uma Thurman is shown walking down •V i l l u s ! stairway and ending up back where she began.

Hpfti \ ,<l tin stairway: Sir Roger Penrose holds the position nl II It. ill Professor of Mathematics at Oxford University. • Penrose and his father, geneticist Lionel S. Penrose

published—for fun—the article 'Impossible f | l | t ' i In A Special Type of Visual Illusion' (British Journal ^Bkyt holog\. vol. 49, 1958, pp. 31-33), describing two •IlKWUll'li objects—the impossible tri-bar and the impossible llffli. ii . M C I scher used the structure of the staircase in his Wi ll known lithograph Ascending and Descending and the


Penrose impossible triangle as the basis of his lithograph Waterfall.

A Mathematician Facilitated and Other Stories: An Indian Journey

Number dealers are everywhere, but they, for some strange reason, did not invade the Indian films as they have done in the West. However, Indian celluloid industry fell in love with mathematics in its infancy. Here are some examples:

• 7 December 1901, Indian film pioneer Harishchandra Bhatvadekar made a newsreel of the public reception given to R.P. Paranjpye who had won a special distinction in mathematics at Cambridge.

• In the Panorama section of the 32nd International Film Festival of India 2001, an English non-feature film titled Vedic Mathematics by K. Jagajivan Ram entered the fray. Veclic Mathematics was adjudged the best scientific film at the 48th National Film Awards.

• One of India's foremost modern playwrights, Girish Karnad was born in 1938 in Maharashtra. His initial schooling was in Marathi. He did his BA at Karnataka University, Dharwad, in mathematics and statistics.

• The song Ande kafunda from the film Jodi No. 1 was written, composed and sung by Pratik Joseph, a young professor of mathematics in a Mumbai college.

• Hrishikesh Mukherjee never even dreamed that he would one day be a part of show business. He was a student of science, and while in college, saw himself as a biochemist studying and teaching flora. After graduating in chemistry, he did teach mathematics and science for a while and then, drawn


I • % Ins I,iscination for photography, signed up with a Kolkata i in In > New Theatres.

I Vnnundlini a Das of the Shakalaka baby and Maithili (Kamala I In . ..in . second wife in Hey Ram) fame has a bachelor's U|||rtT in stalistics and mathematics.

'iQhitlons

| I I Wti.it II It Did Not Win an Oscar

W"inli-i the director was puzzled, he got two numbers 803 and H O Mm- is I solution by Dr R. Raghavendran: L I I LLU IIIIIIIIHT of awards be ABC. Then,

Al»< 711 = A2 + B2 + C2

Ii I .* I .. understood that \ i H' + C2 = ABC/11 < 999/11 < 90

)t|lWi •• 111• I\ llu- following cases:

I HW I

I. h A 9. we have II . C2 90 -81 = 9

llliin' Ihrec digit numbers that start with 9 and are divisible by 11 Hn 'Ml' 'Ml. 924. 935, 946, 957, 968, 979 and 990. However, none

»f Hi. i numbers lit the equation.

i Mr II

I hi A 8, we have II i C2 90-64 = 26

Bgiilii. lluisi- three-digit numbers that start with 8 and are divisible II wr HIH. 814, 825, 836, 847, 858, 869, 880 and 891. Wliil< .ill utliers are inadmissible, 803 is a possible solution, since

11 (K' + 02 + 32) = 11 (64 + 9) = 11 x 73 = 803


Case III

For A = 7, B2 + C2 90-49 = 41

Of all three-digit numbers starting with 7 and divisible by 11 (which are 704, 715, 726, 737, 748, 759, 770, 781 and 792), none is the solution.

Case IV

For A = 6, B2 + C2 90- 36 = 54

605, 616, 627, 638, 649, 660, 671, 682 and 693, are all such three digit numbers that start with 6 and are divisible by 11. However, none of these proves our equation.

Case V

For A = 5, B2 + C2 90-25 = 65

Three-digit numbers starting with 5 and divisible by 11 are: 506. 517, 528, 539, 550, 561, 572, 583 and 594.

While all the other numbers are inadmissible, 550 is anothei possible solution, for

1 1(52 + 52 + 02) = 11 (25 + 25) = 1 1 x 50 = 550

None of the other cases, i.e. A < 5, yield any other solution.

Dear Arnie's Reversible Logic

The two numbers that Schwarzenegger got were 1089 and 9801 Here's how: Let the number be a ^ a ^ . Going by the problem,

= x2, and v w i = y2;


* It* n

m V i ^ 1 , ) , a i + ' ° 2 a 2 + 1 0 ' a 3 + 10°a4 1 a i; a4 9 ;

II II , U, 9

l 'inli i (ilven conditions: HMD x £ 9 9 9 9

III lIlHl I } W

i iinlllrly. I) y ')')

Mm. •*' • V -11(91 (a, +a4)+ I O ^ + j^)} => 11, which is a factor Hfft •

I'uivr Ihnl the above implies 11 is a factor of x and also of y and ^ ^ M t l i h two-digit multiple of 11 satisfies all the conditions and

fw I h'll A Puzzling Look

• M T L R IK I I map-maker.

Cabs, Chains and Birthdays Games Galore

For many of us, it all started with Srinivasa Ramanujan's famous story on a taxicab number plate. If you are not familial with it, no problem, we can come back to that story later. Meanwhile, let me share my story of a different number plale, The number plate of my car reads WB 02L 1905.

How significant is the number 1905? To a science researcher, this may mean the year in which Albert Einstein published Ins special theory of relativity; to a film buff this was the year in which Greta Garbo was born and so on. How you look at n number is important.

I look at 1905 this way. If you add all the digits of 1905, we get 15. Add the digits of 15 and you get 6—a perfect number What is a perfect number? It is an integer that equals the sum of its proper divisors. And what are the proper divisors of 6? 1, 2 and 3. If we add them, we get, 1 + 2 + 3 = 6 . What is the next perfect number? It is 28. How do you know that a number is perfect? Many years ago, German mathematician Leonard Eulercame up with an elegant formula for that. Any even, positive number thai obeys this formula is perfect. The Euler formula is: 2"" '(2" - I)

All you have to do is to put n = 1, 2, 3. . . 100... and start generating your own set of even, perfect numbers.

i . i h s , C h a i n s a n d B i r t h d a y s 23

| | ynii have warmed up, we can get back to the story of the I mill (•' IIIIINI

Mm Qtnlui of Ramanujan

111!* tin idenl has become a part of mathematics folklore, H¥t»lv inr British mathematician G.H. Hardy and Srinivasa

i«<tiiiiimi|in. f)li' i\ while tiavelling in a taxicab from London to Putney,

mi.»• 11• i>• Ramanujan, Hardy noticed its number—1729. D|m..M have thought about it a little because he entered the I mill v I" i' Kamanujan lay in bed and, with scarcely a hello,

Ipflliliiiii his disappointment with it. It was, he declared, 'rather MMlbcr', adding that he hoped that wasn't a bad omen.

M o . I lmd\. said Ramanujan. 'It is a very interesting number. H i ilii minillest number expressible as the sum of two (positive) ^ H i In twn different ways.'

^ B I I I ynii tell me the numbers Ramanujan had in mind?

H ^ l f i I a xi( ,tb Number: A Sequel II you arv weak-hearted)

^ ^ H |h<* Knmnnujan-Hardy story broke, mathematicians came H p l l h n i inn ept called 'taxicab number' . They wanted to

PQVi'i all such numbers that obey a particular rule.

lite Mlh taxicab number is the least integer that can be ^HMlWd us ii smn of two positive cubes in (at least) n distinct |j(H| I stance, the Ramanujan-Hardy taxicab number is

D | | | I I I I I I \ M I as laxicab(2).

I I* In* tuxu ab number is trivially: l.ni. nb( I) = 2

= 1 3 + 1 3


In 1957, using the computer search, Leech discovered Taxicab(3) = 87539319

= 1673 + 4363

= 2283 + 4233

= 2553 + 4143

and in 1991, E. Rosenstiel, J. A. DardisandC. R. Rosenstiel found Taxicab(4) = 6963472309248

= 2421 3 + 190833

= 54363 + 189483

= 102003 + 180723

= 133223 + 166303

In November 1997, the fifth taxicab number was discovered hy David W. Wilson

Taxicab(5)= 48988659276962496 = 387873 + 3657573

= 1078393 + 3627533

= 2052923 + 3429523

= 2214243 + 3365883

= 2315183 + 3319543

Why don't you discover Taxicab(6), Taxicab(7), Taxicab(8) J

Did You Know Who discovered that 1729 could be expressed as a sum of two positive cubes in two distinct ways?

It was not Ramanujan but Frenicle de Bessy in 1657.

Number Chains

All that a mathematician does is to look for numbers every whcm so that s/he can play with them. Here is a simple game thai everyone can play:

C a b s , C h a i n s a n d B i r t h d a y s 25

Dl>l< l Wliu h jersey number did Pele make famous in football?

Kit•|i ' Add to that the number of people who landed on the Moon IMI thr lirst time in human history.

I Multiply the total you got by the number of heroines in tin* lilm \mar Akbar Anthony.

||9p -I II you have followed the steps correctly you must have Hiii .i mi ii nl h I that is a square of a natural number. What is it?

|»|t 11 Revisited

Iwiumi i ill myself acouch potato. However, I became one in ttf hiii ((UiUli i1 of 2001.1 could not believe my eyes when I saw flHM .a u pictures. Here I go again on the World Trade Qflili i ,ni.i. k I know you have heard this many times, and I j j t t mil p n n r repeat it. This is not the book to talk about 0| | | l i ln v hut I aw a number that reminded me of an old school


multiplication trick. How high were the twin towers? 110-storey each. Good, that is our starting point. The number—that is 110—is a product of 11 and 10. All I'll be talking about here is a multiplication trick with 11. If you only knew addition and your teacher asked you to multiply 36 and 11, how would go about it? You can add eleven 36 times and get the answer, but that is a long and cumbersome process. Here is how I would do it—separate 3 and 6 to create space for another number, add 3 and 6 and put it in the middle. The number you get is your answer.

Step 1: 3_6 Step 2: 3 + 6 = 9 Step 3: 396

So, 36 x 11 = 396

Your Turn Does the trick work if you multiply 36 by 111 ? If yes, how?

The Other Game—My Birthday Number

My wife has a standard complaint—our seven-year-old son is not interested in mathematics. One is sure she is not the only mother who has been worried over the math-phobia of her child. Here is a little game all mothers can play with their children who love Popeye more than FI (pi). Let us call it 'My Birthday Number' game. Before we get into the details of the game, here is a story of my birthday number.

I was born on an eighth of a month forty summers ago. Before 1 even started counting, '8 ' became my friend. I was not aware-it appears on the number line not immediately after 7 because many rational and irrational numbers jam the space between them. My fascination with the 'birthday number' grew after

( . i b s , C h a i n s a n d B i r t h d a y s 31

•WHueii v entered my classroom. I was awestruck when I learnt |||MI evei \ sugar cube has 8 vertices. It is also the sum of two MptiiM Vim can express 8 as a sum of two primes (8 = 5 + 3); HI Homier. < ioldbach told us that long ago. By the way, the iMllllH'i 40 is 5 times my birthday number,

i Now we are all set to play the game. Here are the rules:

f in Iwn, I hree. . . or More and an Administrator

1 8 14 20 26

t 9 15 21 27

•1 7 10 13 16 19 22 25 28 31

J 11 17 23 29

ft 12 18 24 30

Figure 4

I I nnk < losely at the grid shown in Figure 4. It has thirty-PViMi' boxes. You start playing from the box where your Mtlrlluliiv number is mentioned.

J*1 All you have to do is touch all other numbers. I n y k uul of movement is allowed. However, for every move

Bfailii one square to another, you have to follow a certain rule. •Ml have to write a mathematical relationship from where

H | H i Hurt to your destination by using their corresponding I Itimihcis and those that appear in between. Remember, you

( I H I I I I I ' I repeat any of these numbers in your equation. ^ H b r tbHtance, if I want to move from, say 1 to 4, then I have to

LMjrilt1 (2 I) + 3 = 4; for my move from 1 to 7, the mathematical • p l u m HI will be 6 + (5 - 4) + (3 - 2) - 1 = 7 and so on.


• Note: Each equation is to be shown to the administrator and only after his/her approval can you move your countel to your destination.

• You are allowed two special chances in the entire game in move to a square of your choice by adding or subtracting u single digit number. However, you cannot use your special chances in succession.

• The idea is to finish the game within a specified time stipulated by the administrator.

• A twist: You can knock off any competitor by predicting the mathematical relation he or she is using to move from one square to the other. However, if your prediction does not match with the one presented to the administrator, you forfeit two chances.

• Time contract: You can announce a time within which you will complete your tour of the birthday circuit. If you fail in do so, you are out of the game.

• Lucky birthday numbers: If your date of birth is a prime number, you get two free moves to adjacent squares. These moves have to be used strategically to get an advantage j

• Prime square: If you land on a prime-numbered square, yotli get a free move. You are allowed this advantage twice and only to adjacent squares. People who have 'lucky birthday numbers', however, are not allowed this move.

• Use of mathematical operation: You can use any mathematical operation as long as all players agree to it.

What Do You Need to Play?

i. A board, as shown in Figure 4 ii. Counters with numbers on them

iii. Writing paper and pencils


I nilni tins

HNC MIL HI K learners, the game can be made easier by altering the I , IIIIIH. ini nlilied game, a mathematical relation can be formed

^ H M H I I M Ihe starting and destination numbers (numbers in-MWft'ii an optional), and using any single-digit number as ^ B Mine* as desired. For example, if the move is to be made

HdR) I in I tin equation can be either of these:

l • I - I "i. I + I + 2 = 4; or, 1 + 3 = 4; or, 1 + 1 + 1 + 1 = 4

fit* Maj'.u i,in in You! iH^MIines birthday parties get boring; the same old tricks, the | | H f iili11iall< ions and the same old cake. You can use numbers

M l loloiit lo your friends' next birthday party. Just do this: ^ • v n i i i age, multiply it by 7, and then multiply that product

liy M n \\ lint do you get? Your age repeats itself three times!

H r t l l i i * 'H (age) x 7 = 196 l % x 1443 = 282828!

)OH II actually doing is multiplying your age by 10101. iNlUg*1 will i. peat four times if you multiply it by 1010101.

i»l Hi..i i Pastime)

lie invented a game at the dead of the night. Sarat i H l I t l l r i 1 -»hout it that he came all the way to meet me at ^ ^ H y v , lie is a nineteen-year-old, young friend of mine who

mi lilnisell with Bach and board game. Here is his gift:

10 flay I'll You Need:

A I' '.Ii I|H d board (see Figure 5) with nineteen slots JL | in. utanil.ird die and a few counters ||| | IH In i Is of paper


How to Play?

A. Players move by throwing the die and start at 0 (not mai keilfl on the board).

B. The objective is to reach slot 19. The first player to rca> li I that slot wins the game.

C. Once you reach a number greater than 9, you do the following! Every two-digit number can be expressed in the form of I lOx + y, so when you cross number 9 and land on any ol llitlII two-digit numbers you move back to x + 2y. For example, I landing on 14 takes you back to 9; for here x = 1 and y 1,1 so x + 2y = 9.

Similarly, for 13 (where x = 1 and y = 3), you go back titH slot 7 (which is 1 + 2*3). This Nova Additum Rule is to I HI applied only once at a time, i.e. if landing on 17 takes you back to 15, you stay there!

Interestingly, landing on 18 takes you back to 17, but landing on 19 leaves you there to be the winner!!

D. Note now that landing on 11, 13 and 17 ONCE is enough


I f||i (lie Nova Additum to take effect. That is, if you land on II lot ih, | II si time you will have to go back to 3, but if it

) H | I | I I I I , lui a second time, you stay there. The same rule ,tirli. lor 13 and 17!

£ h ' i nil "lin t two-digit numbers, except 19, you have to go l»t> I I W K I For example, landing on 12 the first (and the

Hgf 'hi l ) llnie lakes you back to 5, but on the third instance, »MI lay there.

H f t f , however, that landing on 16 takes you to 13, and if it is Mi l l In .1 time there, it is considered as a drop on 13 and so

» i'ii iiny there.

» Ul„ , Null",

IB ) I" pieleiably a three- to five-player game.

Vflteel nl paper can be used to keep track of how many Htlfft 1 pl-iyei has fallen on which number.

fume was invented while investigating 'positional digital •

Hfcif Shkii Km I, Kolkata)

f lbdr ( hains

|» tin Hiiliiiion. li is good to spend a whole afternoon playing the

|i Thf |ritey number Pele made famous in football. I Ailil i" ilittl tlie number of people who landed on the Moon for


the first time in human history. (Two—Neil Armstrong and Edwin 'Buzz' Aldrin)

1 2

Step 3: Multiply the total by the number of heroines in the film Anuu Akbar Anthony. (Three—Shabana Azmi, Parveen Babi and Neelu Singh)

3 6

Step 4: The square root of the above, and the answer, is:

6

Try using different facts and chain them with numbers.

Your Turn

Yes, the trick works for any number multiplied by 111. In fact, il works for any number multiplied by 1111 as well.

For 36 x 111, here is my way of doing it:

Step 1: 3 6 Step 2: 3 + 6 = 9 and place it in front of 6 Step 3:3 + 6 = 9 and place it after 3 Step 4: 3996

So,36 x 11 = 3996

Try it with 99 and 1111 and use the carrying addition rule. Have fun!

Judy's Logic I

| | ) ( f M twenty wonderful logic puzzles created by Judith l|)t(ii<i' Ml Hutchinson, Kansas, America. If you enjoy these • l l i f t , ynii can go over to her site, Judy's Logic Problems. All

ttltonl Judy is her work on the Internet, her multiple pets, H p f t t m s and her love for whodunits.

AfUti|iiPi<tdlng for the Gentry

•Hon: Mr Boddy has invited the lovely (and wealthy) Melon to dine, and now all he has to do is set the stage. It

Ihitl Mi Boddy has made some very unwise investment fP l lo ic . , uiiil as a result, he has had to let his extensive personal ^ ^ • f e t However, after much wheedling and a bit of blackmail, W>rmiiuK d live of the six suspects in the game of Clue to help Mftlinn it nl. Mustard could not be reached as he was on safari finiV"» Master impersonators, the five suspects—Mr Green, ^ K | * m l c i . Mrs Peacock, Mrs White, and Prof. Plum—slipped ^Hapk i ious ly into the roles of butler, chauffeur, cook, downstairs HMIil, uml I ' . i i d e n e r , with each suspect playing a different role. f|M It niiiM|iierader arrived at the Boddy mansion driving a car H | | i r h ' i . - i i t colour—blue, green, purple, red, or white—and MlMi'i)"' i drove a car associated with his or her name. (Note:


Miss Scarlet 's name is associated with the colour red, Mrs Peacock with blue, and Prof. Plum with purple.)

Can you determine, for each suspect, what colour of c d he or she drove, and which member of the staff he or slio impersonated? 1. Prof. Plum didn't drive the red car. 2. Mr Green played the part of the cook. 3. Mrs Peacock, who arrived in the green car, did not masquerade*

as the gardener. 4. Mrs White (who didn't drive a purple car) impersonated

the chauffeur. «

5. The person who arrived in the white car (who wasn't Prof, Plum) played the part of the butler to perfection.

Table 1 POSITIONS CAR COLOUR

BU

TLER

CH

AU

FFE

UR

CO

OK

DO

WN

STA

IRS

MA

ID

GA

RD

ENER

BL

UE

GR

EE

N

PUR

PLE

RED

WH

ITE

MR GREEN GO h MISS SCARLET LU a. MRS PEACOCK (S> 3 C/5 MRS WHITE (S> 3 C/5

PROF. PLUM

OS BLUE

o GREEN o r > PURPLE OS <f RED U WHITE

J u d y ' s L o g i c I 37

mm* Party

o lion The six suspects in the board game, Clue (Col. • M i l l , Miss Scarlet, Mr Green, Mrs Peacock, Mrs White, and

M 1*1 tit11 >. and the six ClueJo characters (Dr Gray, Inspector | ( l ud\ Mrlon, Miss Lavender, Mrs Bluebell, and Sir Sable)

,1 ii 11, illoween costume party last Saturday. Each of them »feHMiip.micd by a date, and each couple went as a famous |i in,ii. -duo. During the evening, a panel of judges determined

• A l n u i ' i ' . of the six monetary prizes (no two of which were | H M I I I ' nun Hint) based on the dance they performed (one couple tllHiftl the rumba) and the costumes worn by them.

H F C L YI HI determine, from the clues given, the man and woman

3III,ii I, up each couple, the costumes they wore (one couple tgll us Ii 'IUI Smith and Pocahontas), the dance they performed lllitl i iii• >• lit the judge's eyes, and the amount of prize that

fill Ii I ullple won?

Mnl ol all the prize awards came to exactly $3,000, and jDpli |let si in's winnings was exactly divisible by 100, although ^H| | |H ip le won exactly $1,000. The second-highest prize

t*linl.'d went to Mrs Peacock and her date (who performed ^ B l n n p i ) Prof. Plum and his date won twice as much as

Mi« Hlui hell and her date, i ^ l f 11 II 11111 • w ho caught the eyes of the panel of j udges duri ng ^ H Willi/ won twice as much as Col. Mustard and his date

•Mfhoie dancing partner wasn't Mrs Bluebell). ioii|ili dressed as Tarzan and Jane won as much as

ftl* While and her date and the couple who caught the eye ill iln MI,I rs during the jitterbug, while Miss Lavender and

lime won as much as Sir Sable and his date and the couple ^ B j l K n l ns Antony and Cleopatra together. (Note: All six ^•Niples are mentioned in this clue).


4. The couple dressed as Hansel and Gretel (who didn't d a m n during the cha-cha) won twice as much as the couple wlwl did a magnificent job of dancing the foxtrot.

5. Dr Gray and his date, and the couple who performed tlitfl foxtrot with perfection, won as much as Miss Scarlet audi her date, and the couple dressed as Lil' Abner and Daisy Maca together.

6. The couple who danced their way into a prize-winning plat d l with their cha-cha (who wasn't Col. Mustard and his datefl won twice as much as Mr Green and his date, who won twicfl as much as the couple dressed as King Tut and Nefertiti.

Table 2

MALE MALE COSTUME

FEMALE FEMALE COSTUME

DANCE PRI/.I: WON

New Year's Resolutions

Introduction: On New Year's Eve, Dr Gray had a few fr iend! over to play Clue. During the course of the evening, it appearcj that each of the six people present had made three resolution! concerning the upcoming new year, one of which was to niakd


Mum i\ |ic of improvement to his or her property. The second tvm illu ii MI concerned engaging in some type of activity that H i Idnl a certain amount of danger and excitement. Last but MHI 11H Ii ast, each decided to either give up or take up something

improve their lives. f r o m the clues, determine the improvement each person

lllk'iid'. in make to his or her property, the activity each plans to |fl|f "l"' in at some point during the year, and what each person K t u l x to either give up or take up in the way of improving the iUNlllv ol his or her life.

Atyu us Improvement: Build a deck, a greenhouse, or a gazebo, trees, put in a patio, or repave the driveway.

4?f'i Itv I >ecp-sea diving, mountain climbing, race car driving, •|f(llvlng. spelunking, and trail hiking.

up/Take up: Give up chocolates or smoking, or take up f i l i n g , reading, travelling, and volunteer work.

H ' t ' l i e lour people who plan to take up something to improve tin i|iiahty of his or her life are Miss Lavender, the person Wlin plans to exercise more (who isn't Dr Gray), the person tylio plans to build a greenhouse, and the person who plans In engage in skydiving some time during the year. Moth ihe woman who plans to put in a patio and the person Win' plans to do some deep-sea diving, plan to either give up Muneiliing or take up something, while Lady Melon plans to tin (lie opposite of what these two have resolved to accomplish.

, I In iv people of the same gender are: the person who plans I In give up smoking (who isn't Mrs Bluebell), the person

M In i plans to do more travelling, and the person who plans In iillenipt mountain climbing that summer.

4 Ni nli. i Inspector Black (who doesn't plan on doing any


trail hiking) nor Sir Sable plan to build a deck on their properties.

5. The person who plans to do more reading, especially of the classics, either plans to do some deep-sea diving or some spelunking some time that summer.

6. Skydiving was definitely not on the list of any future activities of the woman who plans to plant trees that spring

7. Of Mrs Bluebell and the person who plans on doing some mountain climbing, one plans to give up a bad habit while the other plans to take up an activity to improve the quality of his or her life.

8. The person who plans to do some spelunking and the woman who plans to have a gazebo built on her property, either both gave up something or both took up something.

9. Two of those present include the person who plans to do more volunteer work in the community and the person who plans to give up eating chocolates (who also plans to participate in car racing that summer at the local speedway).

Table 3 CHARACTER PROPERTY

IMPROVEMENT ACTIVITY GIVE UP/

TAKE UP


A lilp to Forget!

A t a / i " linn This past winter, the three female Clue suspects M (lie three female ClueJo characters had occasion to fly to mip'pt Although the name of the airline and their destinations Nli'it in it important to this problem, the fact that they had to layover l l ' l (Jlllerent country other than their final destination is.

I luniii' their flight, each woman had the unfortunate luck »»n In Mile an individual who had an annoying habit which

MHiliiitn il throughout the flight until each woman's plane made Milium luiluled landing in one of six different countries—Austria, ^ h f , Greece, Norway, Scotland, or Spain. Finally, they were | N f a l l way, but like it is said, things come in threes, and sure fpui l 'h upon reaching their final destinations, each woman • | n « l that her luggage had apparently been miss-routed and it B M number of days for the luggage to be located and reunited H i t II*• owner. The good news is that the trip back home was • h o o t any incident! From the clues given here, determine the Dftm l"l' habit of each woman's flight companion, the country MlMm h their flights made an unscheduled landing, and the INN*, i 'I i lays it took for their luggage to be located and returned in them

• T h e nix women are: Lady Melon, the woman who had to lnyovei m Austria, the woman who had to layover in Greece, llii woman who sat beside a man who continually cracked |iii< knuckles the entire flight, the woman who sat beside a

1 limn who slept and snored the entire flight, and the woman N i h i l waited a total of four days for her luggage. , The woman whose flight made an unscheduled landing in

N< oll.md had endured the nerve-wracking popping of her i iimpiuuon's chewing gum.

j| Mi White, whose plane made an unscheduled landing in


Norway, had to wait longer to receive her luggage than did the woman who sat beside a man on the plane who spent hours crossing and uncrossing his legs.

4. Mrs Peacock (who wasn't the woman whose plane landed in Greece) wasn't the woman who sat beside a gentleman who constantly pulled on his ear lobe, and who had to wail for three days for her luggage to finally arrive.

5. The total number of days it took the six women's luggage to finally reach them was 22. Together, Mrs Bluebell and the woman who had to layover in Austria had to wait an odd number of days.

6. The woman who sat beside a man who mumbled to himsell throughout the flight, and a woman whose luggage took five days to finally reach her at her final destination, are Miss Lavender and the woman whose flight had to layovef in Spain, in some order.

7. Miss Scarlet, who isn't the woman who sat beside the man who cracked his knuckles, had to wait longer than three days for her luggage to finally reach her.

Table 4 CHARACTER ANNOYING

HABIT DESTINATION DAYS TO GE'I

LUGGAGE

J u d y ' s L o g i c 1 41

Mi noddy's Carousel

MA* iii in I Hiring the last week of October, Mr Boddy invited ityfftil hi" ml over to celebrate Halloween. One of Mr Boddy's

m qulsitions was a carousel with eight horses, each horse a P f t nii i ili mi Each of the eight guests showed up in costume,

U l t i i l i i g ihe course of the day, they enjoyed a ride on the ( I I O I S C S are lettered A through H in the illustration).

H i ' l l guest's surname is associated with the colour of one H | fight horses, although no person rode the horse whose ^HlUkk associated with his or her surname: Col. Mustard—

j^Hty, I )r Gray grey, Lady Melon—orange, Mr Green—green, H I HVendei lavender, Miss Scarlet—red, Mrs Peacock— MM, mill Su Sable—brown.

Mi l you determine, from the given clues, the colour of the • 8 In em 11 position, the guest who rode each horse, and the • P ^ f worn by each guest? (Note: The carousel moves in a H t y l i e position, as seen in the diagram.)

^ H t ill the male guests wore the costumes of a sheikh and |)|filiii k ilolmes, and two of the women guests wore the

^ • I t l i n c s of a witch and Cleopatra. ied horse was directly in front of the horse Miss Scarlet

^ H 0 t t ' H u s was the only instance in which a person rode a | H N next to a horse whose colour was associated with his "i hi i name.

ens horse was directly across the carousel from the H i lioise. and the horse ridden by the man in the Dracula • M i u n e was directly across the carousel from the horse

^ B t l f l l h\ the man who wore the clown costume, ^ p i l ul the guests were Dr Gray (who rode a horse directly

H | U i lioni Ihe red horse), Lady Melon (who rode the horse ^ H p t l y across from the green horse), Mr Green (who rode the


horse directly across from the horse ridden by the guest dressed as a ghost), and Mrs Peacock (who rode the horse directly across from the horse ridden by the guest dressed as a witch l,

5. Sir Sable rode the horse directly across the carousel froil the lavender horse. Col. Mustard rode the horse directly across the carousel from the orange horse.

6. The woman who came dressed as a leprechaun rode thd horse in position H. The brown horse was in position I > Miss Lavender rode the horse in position C. The person dressed as Dracula was not the guest who rode the horse l l position E.

7. Miss Lavender rode the horse which was immediately between the green horse and the horse ridden by the gucil dressed as a witch (who didn't ride a horse immediately ncn| to the horse ridden by the guest dressed as Dracula).

Figure 6

J u d y ' s L o g i c II 43

VlillliiK America's Natural Wonders

MlN/"< nun: One weekend in early fall, the six Clue suspects g^liiliiihci to share their experiences and show slides of some Iff HH' natural wonders each had visited during the summer j ^ B t i l t I iach suspect had visited three different popular vacation ^ • i t n d between them all, they spent some time at six different

III America—the Badlands, Crater Lake, Devil's Tower, ^ ^ • I t t d c s . Mammoth Cave, or the Petrified Forest. No two !Hl(h Itlu.ils visited the exact same three places, and each natural IMHII'I «-IS visited by exactly three individuals.

h u m the clues provided, determine the three different natural ^B |VI<I each person visited during his/her summer vacation.

, f f W o l llie three individuals who spent some time exploring Mil* Morula Everglades also visited the Petrified Forest in AIL/ona.

| Mlween Col. Mustard and Mr Green, they visited all six H| | t l i rul wonders.

Ifiiiih Dakota's spectacular Badlands was visited by both • M r < Irccn and Mrs White. Although neither Mr Green nor

MR. White spent any time at Crater Lake in Oregon, both ' {VI Mustard and Miss Scarlet did. • M l < irccn visited exactly two natural formations which Tweie also visited by Miss Scarlet, and Mrs White visited " 91 in lis one natural formation which was also visited by Miss

Vnr lc t .

Ii Nuih < ol. Mustard and Prof. Plum showed slides of their •Vlkll to Mammoth Cave in Kentucky. Neither of these two Hylklletl the Petrified Forest, although both Miss Scarlet and

Mi I't acock did. f I Km I \ l ower, in Wyoming, was visited by both Mrs Peacock

mill Mis White.


7. Prof. Plum ended up spending several more days than originally intended in the Everglades, as he found thciflj utterly fascinating.

Table 5

CHARACTER FIRST NATURAL WONDER

SECOND NATURAL WONDER

THIRD NATURAL WONDER

Halloween Treats

Introduction: This year, during the last weekend of October, fivtfl ClueJo characters—Dr Gray, Lady Melon, Miss Lavender, Mm I Bluebell and Sir Sable—were in America visiting friends, eacM of who is one of the Clue suspects—Col. Mustard, Miss Scarlet J Mrs Peacock, Mr Green, or Prof. Plum. Each ClueJo character was the guest of one of the Clue suspects. In America, Halloween falls on the last day of October, a night when youngsters gnl out in costumes to 'trick-or-treat' at homes around the arcaJ Each of the ClueJo characters enjoyed helping his or her host! or hostess in handing out Halloween treats to the children. All


home, from one to five different types of treats were PHIInhle apples, bubble gums, candy bars, cookies, and/or H^ly Inn s No two households gave out the same number of feBpi nl Heats, although each treat was given out by exactly three glltmini households.

I'loni ihe clues below, determine the name of each host or Hmnm the name of his or her guest, and the treat or treats each

•H* mil lo (rick-or-treaters on Halloween night.

|*iol Plum and his guest and Mrs Bluebell and her host or l |||Hitess did not give out any of the same types of treats to

Mm their little costumed visitors, i {Miss Scarlet and her guest handed out one type of treat that fcJUfli'' also handed out at the homes where both Lady Melon

mill Miss Lavender were guests. \i Mi Peacock's home, trick-or-treaters either got both

Htytplcs and cookies, or they got neither, f j ml\ Melon was the guest of the person who gave out at H M * I I wo different types of treats, neither of which was

bubble gums or apples. Cnl Mustard and his guest did not hand out all five types nl Heals, although he and his guest, along with Prof. Plum

• m l lus guest, did hand out popcorn balls. No popcorn balls fltyeic given out at Mr Green's home. H t p l u e e n the home where DrGray was a guest and Mrs L l f i u in k s home, all five types of treats were given out, • l l lhoi i fh neither household gave out the same type of treats. , Al Ihe homes of Col. Mustard and the home of Mr Green

^Hritosc guest was not Miss Lavender), either both gave out H|||i|>lcs or both gave out cookies, but in only one case (if at

nil) wen- both apples and cookies given as treats.


Table 6 HOST/HOSTESS GUEST TREAT/S

A Restful Retreat?

Introduction: One weekend at the end of September, following an especially hot summer. Miss Scarlet and the other five Clud suspects decided to spend some time relaxing at the Rainbod Retreat in Echo Valley. By chance, each of them was given one of the first six cabins—three of which are on the north side of Rainbow Lane (west to east) numbered 1,3, and 5, and threJ on the south side of Rainbow Lane (west to east) numbered .'J 4, and 6 (see the diagram). Number 1 is directly across front number 2, etc. as illustrated in the diagram. Each stayed in u cabin of a different colour—lavender, lemon yellow, lime green J peach, pink or sky blue.

The first night was anything but restful, as each was k<pl awake most of the night by a different sound in or near his or h e r cabin—barking dog, buzzing fly, chirping cricket, croaking frog, hooting owl, or yowling cats. From the clues given here, determine the number and colour of each person's cabin and the sound which kept him or her up all night.


If filliping crickets kept the person up who had the cabin kjittiediiiicly between Mrs Whites' and the person who had |||i« leiuim-yellow cabin.

• M Peacock's cabin was directly across from Mr Green's lihlu hiuI next to that of the person who tried in vain to kill

• | l U / / i i i g lly in his or her cabin. ( ^ h ' lime gieen cabin was on the same side as, and somewhere

east of. Prof. Plum's cabin. i J f l l i '-k\ blue cabin was south of, although not necessarily ItUN* Hy -'cross from, Col. Mustard's cabin. < | In number of the cabin of the individual who was unable

I nice11 because of croaking frogs (which was somewhere .,i i<t ihe peach cabin) was exactly twice that of the iVldual who listened to cats yowling all night long.

• | t ( i pink cabin had a smaller number than Mrs Peacock's H | h t n and a larger number than the lavender cabin. H u l Mustard's cabin number was more than twice that of H | # person who was kept up by a hooting owl.

West

Figure 7

3 5

Rainbow Lane Rainbow Lane

4 6

• iu» qu i l t

| | V l f t / " < " ' M r s w h i t e k n e w t h a t D r Gray's sixty-first birthday ^ M i m i n g up, and since Clue is his favourite game, she decided

lie him a quilt appliqued with symbols from the Clue game.


The quilt was to have eight symbols, two each of the candlestick J knife, lead pipe, revolver, rope, and wrench, as well as two each of the green token (the one Dr Gray prefers to use when he play* J and two each with the word 'Clue'. She placed the squares with the appliques in no particular order on the quilt. From the clued given, can you determine which two squares on the quilt (lettered! A through P in the diagram) are matching? Note: 'Left' and 'rightl refer to the solvers' left and right as they look at the diagram. I

1. The squares with the candlestick are A and K. The appliqudl in square C matches that in square J, and the appliqud in square F matches that in square M.

2. Mrs White got so carried away trying to place the squalen depicting the weapons, she didn't realize until all was sewn together that she had placed the two squares with the word 'Clue' next to one another, with one directly above the othet '

3. The two squares depicting the wrench were diagonal from one another, touching in the corner. One is to the immediate left of a square depicting the lead pipe, and the other is to the immediately left of a square depicting the revolver.


l | f lot It squares depicting the knife are in corner positions. One Silt the squares with the knife applique is directly below the

In* m which a square depicting the green token can be found, • l i t r e consecutive squares in one row, from left to right, are ljit|liiiics appliqued with the revolver, the rope and the green

lukclt.

olyllons

Bfc|tirt<iding for the Gentry

H f c id accompanying the problem, use an X for a 'no' answer and ftil II yes' answer. By going across the grid, one can put an X in

^Mt|iiiiie that intersects Mr Green and green under the Car Colour • f t l i . an X in the square that intersects Miss Scarlet and the red, and

liei ausc no person drove a car which was a colour associated lie. HI her name. Since, according to the given clues, Mr Green

the cook, Mrs White the chauffeur, and Mrs Peacock drove the Bill i at then it was Miss Scarlet who drove the white car and played

H|)Utlci Next, since Prof. Plum didn't drive the red car or the purple dime the blue car. We also know Mrs White didn't drive the

H f c SHr, so she drove the red car, and by elimination, Mr Green M l tin purple car. Lastly, Mrs Peacock did not play the gardener, so • nliiyi d the downstairs maid, and by elimination, it was Prof. Plum

Hnpliivi'd the gardener.

^•ynmiiiy: (suspect, colour of car, member of staff) l | 4 r < Irccn, purple, cook i M l i Peacock, green, downstairs maid P M I I While, red, chauffeur BM | , , s Scurlet, white, butler

I'm! I'lum. blue, gardener


Costume Party

The six Clue suspects are Col. Mustard, Miss Scarlet, Mr Green, I Mrs Peacock, Mrs White and Prof. Plum, and ClueJo characters an I Dr Gray, Inspector Black, Lady Melon, Miss Lavender, Mrs Bluebell I and Sir Sable. Given that the prize money totals $3,000, and that though I each prize amount was divisible by 100, no one won exactly $ 1 ,CXX), I the possible amounts are $100, $200, $300, $400, $500, $600, $7(K), I $800 and $900.

We know that the one who won the first prize (highest amount)I danced the cha-cha and that the second-highest prize award went to I Mrs Peacock and her partner who danced the tango. Also, from clue 3, I the couple dressed as Tarzan and Jane won as much as Mrs White anil I her partner and the couple who did the jitterbug together, anil I Miss Lavender and her date won as much as Sir Sable and his partner I and the couple who came as Antony and Cleopatra together; so the I couple who dressed as Tarzan and Jane and Miss Lavender and her I partner were the ones who won the first and second prizes. Thus, Miss I Lavender and her partner won the first prize and danced the cha-cha. I and Mrs Peacock and her partner who danced the tango came dressed I as Tarzan and Jane.

Now, Miss Lavender and her partner who danced the cha-cha and came first won twice as much as Mr Green and his date, who won twice as much as the couple dressed as King Tut and Nefertiti, and it Miss Lavender and her partner won $900, then Mr Green and his partner won $450, which is impossible (not divisible by 100), and the same is true if they won $700. So, Miss Lavender and her partner won the first prize of $800, Mr Green and his partner won $400, and the couple I who came as King Tut and Nefertiti won $200. Since Miss Lavendei I and her partner won as much as Sir Sable and his partner and the couple I dressed as Antony and Cleopatra, the total amount won between the I latter two couples comes to $ 1600, leaving $ 1400 for the couple dressed as Tarzan and Jane, Mrs White and her partner, and the couple who danced the jitterbug (clue 3), and the total for the couple who won the second place could only be $700, which total between Mrs White


A l I" 1 partner and the couple who danced the jitterbug totalling $700. H L Clue 6, we know the three totals were $800, $400 and $200, Kill Mis I'eacock and her partner won $700 for a total of $2100, so • j l Wlule and her partner, and the couple who did the jitterbug won MNnI i >1 VMX) between them ($3000 total, inferred from clue 1), but it flft he $ 7(K> and $100 (the couple who came second dressed as Tarzan Hd Iiiue won $700) nor can it be $500 and $400 (each couple won a Mpipui amount, and since four prizes were $800, $400, $200 and Ifftll. then $500 and $400 are impossible). H bus. Mrs White and her partner and the couple who danced the

mg, won $600 and $300, in some order, and the six prizes are m, 1. /DO, $600, $400, $300 and $200 (amounting to $3000).

PlAl Ihis point, we have at the first place Miss Lavender and her • ( M i dancing the cha-cha and winning $800; at the second place is

peacock and her partner dressed as Tarzan and Jane, dancing the Mtylu int.I winning $700; the third place winners get $600; at the fourth |l|||ii' i Mr Green and his partner winning $400; the fifth place winners

H|,1(N). and the sixth place winners, dressed as King Tut and Nefertiti, • | ? ( H l Thus, from clue 3, since Miss Lavender and her partner (first plfr i i lin cha) won $800, then Sir Sable and his partner and the |Hl|i|i' ili essed as Antony and Cleopatra were the ones who won $600

H | | 20 l ) . and since the sixth place winners of $200dressed as King Tut S j | | Nelcrtiti, then that was Sir Sable and his partner, and the third

mners of $600 dressed as Antony and Cleopatra. The couple llltnccd the waltz won twice as much as Col. Mustard and his

Euei and since Mr Green and his partner won $400 and Sir Sable Ills partner (King Tut and Nefertiti) won $200, and Col. Mustard

Ulilu'i have won $800, $700, or $600 (not amounts twice those), HI ('nl Mustard and his partner won $300 and got the fifth place, | the couple who danced the waltz won the third place of $600 as liny and Cleopatra.

I The couple dressed as Tarzan and Jane won as much as Mrs White Itci partner and the couple who danced the jitterbug; so Mrs White

Kill liei partner and the couple who danced the jitterbug were fourth

100 M i n d s t r e t c h J u d y ' s L o g i c II

52

(Mr Green and his partner, $400) and fifth (Col. Mustard, $300), in some order. The couple dressed as Hansel and Gretel won twice all much as the couple who did the foxtrot, but they didn't dance the cha-cha, and hence weren't first—so it could only be Mr Green and his partner (fourth. $400) who dressed as Hansel and Gretel, and the couple who did the foxtrot was Sir Sable and his partner (sixth, $2(X)), Dr Gray and his partner and the couple who won with the foxtrot, won as much as Miss Scarlet and her date and the couple dressed as Lit' Abner and Daisy Mae; since the second-place winners dressed as Tarzan and Jane, and the third-place winners dressed as Antony and Cleopatra, and the fourth-place winners (Mr Green and his partner, $400) dressed as Hansel and Gretel, then, the couple dressed as Lit' Abner and Daisy Mae can't be Miss Lavender and her partner (first, cha-cha, $800), as $800 added to any other total would be more and not equal to any won by Dr Gray and the couple who danced the foxtrot

Thus, it was the fifth-place winners (Col. Mustard and his partner, $300) who dressed as Lil' Abner and Daisy Mae, and by elimination, Miss Lavender and her partner dressed as John Smith and Pocahontas Miss Scarlet wasn't Sir Sable's or Col. Mustard's partner, so she was either Mr Green's partner, or she and her partner came as Antony ami Cleopatra. If she was Mr Green's partner and they won $400 and added to the $300 which Col. Mustard and his partner won as Lil' Abner and Daisy Mae, then it totals $700; so Dr Gray and his partner would have won $500 to add to the amount won by the couple who danced the foxtrot (Sir Sable), but we've determined that no couple won $500; therefore, Miss Scarlet won $600 and so those winnings plus Col. Mustard's of $300 equal $900. Thus, Dr Gray's winnings added to that of the couple who danced the foxtrot must equal $900, so Dr Gray can only be Mrs Peacock's partner.

Prof. Plum and his partner won twice as much as Mrs Bluebell and her partner and since Dr Gray won $700, then Prof. Plum was either Miss Lavender's partner or Miss Scarlet's partner, but he can't have won $600, as Mrs Bluebell would have won $300 and been Col. Mustard's partner, which she isn't. Thus, Prof. Plum was Miss Lavender's partner

MI» bluebell was Mr Green's partner, and by elimination, Miss IfWW' partner was Inspector Black.

Mr* White wasn't Sir Sable's partner, so she was Col. Mustard's ,Mtttll»ii mid the couple who danced the jitterbug was Mrs Bluebell • M r < Irccn. By elimination, Mrs White and Col. Mustard danced tityhliiiki, and Sir Sable's partner was Lady Melon. ^Plfltitry: (the couple, costumes, dance, prize amount) S t l Mustard and Mrs White, Lil' Abner and Daisy Mae, rumba,

• f d m y and Mrs Peacock, Tarzan and Jane, tango, $700 |fe|r< t. ii Black and Miss Scarlet, Antony and Cleopatra, waltz, $600

• H l i c c n and Mrs Bluebell, Hansel and Gretel, jitterbug, $400 Mil hum and Miss Lavender, John Smith and Pocahontas, cha-

,„, IK.*)

fill Suhlr and Lady Melon, King Tut and Nefertiti, foxtrot, $200

| Year's Resolutions <il all the characters involved are Miss Lavender, Dr Gray,

Melon. Inspector Black, Sir Sable and Mrs Bluebell. While four plan to take up something to improve the quality of their life,

n plan to give up something. The women plan to put in a patio, I'I X and build a gazebo, and the men plan to build a deck, build (muse, and put in a new driveway. The person who plans to

a di i k isn't Inspector Black or Sir Sable, so it is Dr Gray. The who plans to do mountain climbing isn't the one who plans up smoking or do more travelling, do more reading (rather

ltl|i i ir spelunking), give up chocolates (car racing), or do more r work (four people of the same gender would be the ones

up i hocolates, giving up smoking, and do more travelling, as ml.I make four people of the same gender, if including one

• plans to climb mountains, which is impossible as there are only I tltfff vMnnrii and three men).

<hc person who plans to do mountain climbing plans to


do more exercises. Since he or she also plans to exercise more, then Mrs Bluebell, who plans to give up something—but not smoking J is the one who planned to give up chocolates and do some car racing, The person who plans to do mountain climbing (more exercise), givJ up smoking, and do more travelling belong to the same gender, hui aren't females, so all are males. Hence, the males are the ones who pliul to give up smoking, travel more, and exercise more, and the femaloJ are the ones who plan to give up chocolates, read more and do mod volunteer work. The man who plans to build the greenhouse isn't thi one who plans to give up smoking or do more exercising, so he is tlJ one who plans to travel more. Since Dr Gray doesn't have any plan til travel more or exercise more, he is the one who plans to give up smoking, and by elimination, the man who plans to exercise more is the onJ who plans to put in a new driveway (inferred from above).

Since Mrs Bluebell is the only female to give up something, Lady Melon must be one of the women who plan to take up something! worthwhile, so Mrs Bluebell must be the woman who plans to put iiJ a patio, and the person who plans to do deep-sea diving can only htfl a male. Since the person who plans to deep-sea diving is a male, audi the person who plans to do more reading is a female, she is the on« who plans to do spelunking, but neither is she the woman who plai J to build a gazebo, nor Mrs Bluebell who plans to put in a patio. Thud the woman who plans to do more reading is the one who plans to plana trees, and so can only be Miss Lavender. By elimination, it is Lady Melon who plans to build a gazebo and do more volunteer work, and sincJ she plans to take up something she is the one who will do skydivingl Therefore, the person who plans to do frail hiking is a male. Since tlJ woman who plans to put in a patio plans to give up chocolate, Dr Gray! would be the person who plans to do deep-sea diving. Inspector Blacjl doesn't plan on trail hiking, so he is the one who plans to exercisJ more, and by elimination, it is Sir Sable who plans to do trail hiking!

In summary: (name, property improvement, activity, give/take up) I Dr Gray, build deck, deep-sea diving, give up smoking Inspector Black, repave driveway, mountain climbing, exercise moJ

J u d y ' s L o g i c 1 55

I ,IIIN Melon, build gazebo, skydiving, more volunteer work Miss I avender, plant trees, spelunking, more reading Mi Bluebell, put in patio, race car driving, give up chocolates Ku Sable, build greenhouse, trail hiking, more travelling

A I lip to Forget! w know ii took one woman five days to get her luggage, and at least • woman's luggage took four days. Since there was a total of twenty-| p i diiy s lor the luggage to be returned between the six women, then

B | Other four women's luggage, between them, took thirteen days. only possible combination of days possible would be for one of

^ • f o i u to have had to wait for four days, and the other three women flf Ihree days each. Thus, the total number of days for the six women

3 4 4 5, in some order. Mile only woman to wait five days for her luggage was either

I ^ B | nvender or the one who had to layover in Spain. Mrs White, ftj|l».r I h(',ht landed in Norway (longer than any another woman), was B o f the two women whose luggage took four days. From the clues •Hf m, it is clear that Mrs White didn't sit beside a person who popped

•MM, I II the man who crossed and uncrossed his legs, or the man who Btyvd on his ear lobe, or mumbled to himself, so she sat beside either H f c i i n who cracked his knuckles or the man who slept and snored. The l | i i woman whose luggage took four days was either Mrs Bluebell BtfWi woman whose flight landed in Austria (has to be three and four I B k lo he an odd number, as 5 + 3 = 8 and 3 + 3 = 6, both even ^•Rhers). but the latter couldn't be one whose luggage took four days

By* 11 Therefore, Mrs Bluebell's luggage took four days to reach her. I f r u n u lue 7, we know. Miss Scarlet's luggage took more than three

'.< i her luggage took five days, and the three women who waited ^ B diiyt each for their luggage were Lady Melon, Miss Lavender

0nl Mi I'eacock. Since Miss Scarlet's luggage took five days, she is f u u r whose flight landed in Spain, and Miss Lavender sat beside

Hinn who mumbled to himself. Since Miss Scarlet didn't sit beside

[


the man who cracked his knuckles, she alone can be the woman who sat beside the man who snored through the entire trip, and that leave* Mrs White as the one who sat beside the man who cracked his knuckles One woman who waited three days for her luggage, sat beside the man who pulled on his ear lobes, but wasn't Mrs Peacock or Miss Lavender I It was Lady Melon who sat beside man who pulled on his ear lobes Since Miss Lavender sat beside the man who mumbled and Lady Melon sat beside the man who pulled on his ear lobes, and Mrs White waited four days for her luggage, then the woman who sat beside the man who crossed and uncrossed legs also waited three days, and by elimination, can only be Mrs Peacock, leaving Mrs Bluebell as one who sat beside the person who popped chewing gum. So, Mrs Bluebell was the one whose flight stopped in Scotland.

Lady Melon's flight didn't stop in either Austria or Greece so il stopped in France. Mrs Peacock's flight didn't stop in Greece, so hct plane stopped in Austria, and the one whose flight stopped in Greece was Miss Lavender.

In summary: (lady, neighbour's habit, layover country, days of waiting) Lady Melon, pulled on his ear lobe, France, three days Miss Lavender, mumbled to self, Greece, three days Miss Scarlet, man snored, Spain, five days Mrs Bluebell, popped gum, Scotland, four days Mrs Peacock, crossed legs, Austria, three days Mrs White, cracked knuckles, Norway, four days

Mr Boddy's Carousel

We know that Miss Lavender rode horse in C, the brown horse waul in D, and the woman who wore the leprechaun costume rode horse in H. Women wore costumes of leprechaun, witch and Cleopatra, and men of sheikh. Sherlock Holmes, Dracula and clown, so the one who wore the ghost costume was a female. We also know that the carousel moves clockwise. Dr Gray rode his horse directly across the red horse, Lady Melon was directly across the green horse, Mr Green directly


In lie.', the ghost and Mrs Peacock directly across the witch. Thus, • Me who wore the ghost costume and the one who wore the witch ^puuic were Miss Lavender and Miss Scarlet, in some order, and ft une who rode the red horse and the one who rode the green horse H i t Sir Sable and Col. Mustard, in some order. Sir Sable rode his IHttM ilneetly across the lavender horse and Col. Mustard rode his horse

across the orange horse (clue 5) which would be Dr Gray Kill I mis Melon, in some order, but Col. Mustard didn't ride the

^ • p l t hi u se across Lady Melon (no one rode a horse whose colour H ) associated with his or her name). So Col. Mustard rode the red

Sir Sable the green horse, Dr Gray the orange horse, and Lady HMtm the lavender horse. Dr Gray rode his horse directly across the ^Bfc>rsc, who was on a horse directly in front of Miss Scarlet. Since ^Hnh le was directly across Lady Melon (from above) and Mr Green ARM* Ihe female who wore the ghost costume (clue 4), then the man ^ f t w i ire the Dracula costume and the man who wore the clown costume | | H ilneetly across one another, were Dr Gray and Col. Mustard,

^H)n i e ol der. Thus, Sir Sable and Mr Green were dressed as a sheikh Ail Sherlock Holmes, in some order. I MI. . I .avender rode her horse between the green horse and the

iidden by the female dressed as a witch (clue 7), and since H k I avender rode the horse in C and the brown horse was in D, • F T Sii Sable, on the green horse, was in B, and the female dressed

H | bitch, in D, was on the brown horse (see illustration). Thus, Lavender wasn't dressed as a witch (witch and ghost were I,avender and Miss Scarlet, some order), so Miss Lavender was

^ ^ M d as the ghost and the woman on the brown horse, in D, dressed ^ • t h r witch was Miss Scarlet. Miss Scarlet (witch, brown horse, D)

IN* diii i ily behind the red horse (clue 2) which was ridden by Col. iWlMiinI (clues 4 and 5) which would be E and directly across Mrs

i H k. and since H is directly across D, then Mrs Peacock was dressed • lltr leprechaun, and by elimination, Lady Melon (lavender horse)

f t l l f ssed as Cleopatra. Miss Lavender (ghost) in C was directly across (l ieen (Sherlock Holmes), so since G is across C, then Mr Green

I


rode the horse in G. Since Col. Mustard rode the horse in E, then Dr Gray on the orange horse was in A. Sir Sable in B (sheikh, green horse) was directly across Lady Melon, so Lady Melon rode the horse in F. Thus far, we've determined that Dr Gray rode the orange horw in position A, Sir Sable was on the green horse in B, Miss Scarlet on the brown horse in D, Col. Mustard on the red horse in E, and Lady Melon on the lavender horse in F; and the grey horse was directly across the blue horse, then those can only be the ones ridden by Mis* Lavender in C and Mr Green in G, in some order, and Mrs Peacock in F must have ridden the yellow horse. Since Mrs Peacock was in F, next to G (Mr Green) and her name is associated with the colour blue, then Mr Green didn't ride the blue horse (clue 2); he rode the grey horse and Miss Lavender the blue horse. Miss Scarlet wasn't immediately next to the guest dressed as Dracula, so it was Dr Gray who was dressed as Dracula, and Col. Mustard in E (next to D, Miss Scarlet) was dressed as the clown (clue 3).

In summary: (horse, person, costume, horse's colour) A, Dr Gray, Dracula, orange B, Sir Sable, sheikh, green C, Miss Lavender, ghost, blue D, Miss Scarlet, witch, brown E, Col. Mustard, clown, red F, Lady Melon, Cleopatra, lavender G, Mr Green, Sherlock Holmes, grey H, Mrs Peacock, leprechaun, yellow

Visiting America's Natural Wonders

Each person visited three different natural wonders, and no two visited the exact same three places (introduction). Mrs Peacock and Mrs Whit J both visited Devil's Tower. Prof. Plum visited the Everglades. Botlt Col. Mustard and Prof. Plum visited Mammoth Cave, but neither visited the Petrified Forest, although Miss Scarlet and Mrs Peacock did. Both i Mr Green and Mrs White visited the Badlands, but neither visited


•pilei I .ake, although Col. Mustard and Miss Scarlet did. Between (VI Mustard and Mr Green, they visited all six places; so since )||f t heen visited the Badlands, then Col. Mustard did not and since |Si| Mustard visited Mammoth Cave, then Mr Green did not, and we gfclw that Mr Green didn't visit Crater Lake, but Col. Mustard did. fi|| r ( ol Mustard didn't visit the Petrified Forest, then Mr Green ^ « l n c c each place was only visited by three people, and the Petrified Hkfml wits visited by Miss Scarlet, Mr Green, and Mrs Peacock, then l^ii'iii I visited by Col. Mustard, Mrs White, or Prof. Plum. Mr Green

H | Miss Scarlet visited exactly two of the same places and one was ll* I'rti died Forest, but the second wasn't Crater Lake, Mammoth Cave

B | « d y visited by Col. Mustard and Prof. Plum, so it would make • n people visiting the same place, and in the introduction it is said H »m h place was visited by exactly three individuals), or Devil's H|kei (same reason as Mammoth Cave), so the second place they Hilled in common was either the Badlands or the Everglades.

• I I both Mr Green and Miss Scarlet visited the Everglades, then ^ • H r c c who visited the Everglades were Mr Green, Miss Scarlet,

Piol Plum. Miss Scarlet then would have visited the Everglades, ^K|ei I ake, and the Petrified Forest (from above) and Mr Green would

visited the Badlands, Everglades, and Petrified Forest. Mrs White B M l s s Scarlet visited exactly one common place, but not Mammoth H (would make four people), the Petrified Forest, Crater Lake • l i t Scurlet and Col. Mustard, so Miss Scarlet would have it in • n i m i n with both Col. Mustard and Mrs White), Badlands (both

I liven and Mrs White, so Mrs White would have had it in common i nli hi itli Mr Green and Miss Scarlet), or Petrified Forest (already

^ H i m n c d that the three who visited it are Miss Scarlet, Mr Green, All Mis Peacock); therefore, Miss Scarlet and Mrs White both visited

• v i l •• lower, along with Mrs Peacock which were the only three to B | Devil's Tower (introduction). Thus, since Mrs White visited the HpHlhls. then Miss Scarlet did not. That would mean that Col. Mustard H| | l r i l the Everglades by elimination (saw both Crater Lake and Huiinii>1 It Cave, but not Petrified Forest, Badlands, and Devil's Tower


by elimination). And since Col. Mustard would have visited Everglades, then Mr Green didn't, but then that leaves Mr Green visiting only two places—Badlands and Petrified Forest, which is impossible. Therefore, of the two that Mr Green and Miss Scarlet both visited in common-J either Badlands or Everglades, it wasn't the Everglades (all th« preceding); they both visited the Badlands.

Since Mrs White visited the Badlands (clue 3), and Miss Scarlrt also did, then that is the only place they had in common, so Miss Scarlet did not visit Devil's Tower—the three places she visited were Badlands, Crater Lake, and Petrified Forest. Two of the three people who visited the Petrified Forest (Miss Scarlet, Mr Green, and Mrs Peacock) also visited the Everglades, but it wasn't Miss Scarlet (out by elimination), so both Mr Green and Mrs Peacock visited the Everglades. Thus, Mi Green visited the Badlands, Everglades, and Petrified Forest, and Mrs Peacock visited Everglades, Petrified Forest, and Devil's Towed and since the Everglades were visited by Mr Green, Mrs Peacock (from above) and Prof. Plum, then they were not visited by Col. Mustard, Miss Scarlet, or Mrs White. Since Col. Mustard didn't visit the Badlands, Everglades, or Petrified Forest and did visit Crater Lake and Mammoth Cave, then the third place he visited was Devil's Tower. By continuing to eliminate back and forth between those facts that are known anil unknown, the puzzle is solved.

In summary: (name, natural wonders visited) Col. Mustard: Mammoth Cave, Crater Lake, Devil's Tower Miss Scarlet: Badlands, Crater Lake, Petrified Forest Mr Green: Everglades, Badlands, Petrified Forest Mrs Peacock: Everglades, Petrified Forest, Devil's Tower Mrs White: Mammoth Cave, Badlands, Devil's Tower Prof. Plum: Mammoth Cave, Everglades, Crater Lake

Halloween Treats

Each household gave out a different number of types of treats, front 1 to 5. The guest at the home where all five treats were given out wasn'l


Mi" llluebell. Lady Melon, Dr Gray, (from clues 1,4 and 6) so was Alk'i Sir Sable or Miss Lavender. If it was Miss Lavender, then she flin t slay with Prof. Plum, Miss Scarlet, Mrs Peacock, Col. Mustard, |f Mi (ireen; thus, it couldn't be Miss Lavender who was a guest at |jt«i home which handed out five different types of treats—it was Sir ftltle Sir Sable didn't stay with Prof. Plum, Mrs Peacock, Col. Mustard d Mi < ireen, so Sir Sable was a guest of Miss Scarlet (all five types). H*li neat was given out by exactly three different households, and

ft (he other two households who gave out popcorn balls were Col. jjtiisi,lid's and Prof. Plum's. Between the home where DrGray was a Burst and Mrs Peacock's home, all five were given out, but neither

ftimcliold gave out the same type of treats—so they were two and ft types, or one and four types.

I Since it was Miss Scarlet, Col. Mustard, and Prof. Plum's households which gave out popcorn balls, then Mrs Peacock's did |Nll thus, Dr Gray had to be the guest of either Col. Mustard or |jjt,l plum (clue 5). Since either Col. Mustard or Mr Green both gave |UI either apples or cookies (or perhaps one gave both), and we know llliiK Scarlet and her guest, Sir Sable, gave out all five types, then ft| Peacock couldn't have given either. Thus, the household where ft, (ii.iy was guest did give out both apples and cookies (clue 6)

H n g with the popcorn balls. The one type of treat that was given Ity Miss Scarlet (Sir Sable), and the homes where Lady Melon and Mi v. I avender were guests wasn't bubble gums or apples, and since

B( i ray stayed with either Col. Mustard or Prof. Plum (popcorn balls), j | |n it wasn't popcorn balls, and since Dr Gray stayed at the home .tyliric both apples and cookies were given out (from above), then it Himn'I cookies; therefore, the one treat given out by households where | | f Sable, Lady Melon, and Miss Lavender were guests gave out candy p s . and since this was the case, then either Lady Melon or Miss Lavender

ftfi to be Mrs Peacock's guest (since Dr Gray couldn't have stayed tyllliperson who did). If Lady Melon stayed with Mrs Peacock, then ||r v didn't give out popcorn balls (Dr Gray and host/hostess did), apples p> kies, or bubble gums, so would have given out only candy bars.


But Lady Melon was a guest of the suspect who gave out at least two types of treats (4), so Lady Melon didn't stay with Mrs Peacock; Miss Lavender was Mrs Peacock's guest. Since Sir Sable and Dr Gray (either Col. Mustard or Prof. Plum) were guests of suspects who gave out apples, then the third household where apples were treats wasn't Mrs Peacock's (Miss Lavender), or Lady Melon, so it was the household where Mrs Bluebell was the guest. Therefore, Prof. Plum and his guesl didn't hand out apples so his guest wasn't Dr Gray (apples, cookies, popcorn balls), so it was Col. Mustard with whom Dr Gray stayed. Since Mrs Bluebell wasn't the guest of Prof. Plum, then she was Mr Green's guest, and by elimination. Lady Melon was Prof. Plum's guest. Since Mrs Bluebell was the guest of the person who gave out apples, then Mr Green gave out apples, and since Col. Mustard (Dr Gray) also did give it along with cookies, then Mr Green did not give out cookies (clue 7), and since cookies were given out by three households, and we know Mrs Peacock didn't give out cookies (Col. Mustard and Dr Gray did), then it was Prof. Plum (Lady Melon) who gave out cookies, along with popcorn balls and candy bars. Bubble gums were not given out by Lady Melon and Prof Plum, so that household only gave out three kinds of treats—candy bars, cookies and popcorn balls. Thus, Col. Mustard had given out four types of treats—three of which we know—apples, cookies and popcorn balls, and Mrs Peacock gave out one type which was candy bars, so Col. Mustard and Dr Gray also gave out candy bars.

By elimination, it was Mr Green and his guest Mrs Bluebell who gave out two types of treats, one of which was apples, and by elimination, the other had to have been bubble gums.

In summary: (host, guest, treats) Col. Mustard, DrGray—apples, bubble gums, cookies and popcorn

balls Mr Green, Mrs Bluebell—apples, bubble gums Miss Scarlet, Sir Sable—apples, bubble gums, candy bars, cookies

and popcorn balls


Mis Peacock, Miss Lavender—candy bars b Prof. Plum, Lady Melon—candy bars, cookies and popcorn balls

• Restful Retreat?

Anl Mustard's cabin was north of the sky-blue one and the number |f his cabin was more than twice that of the person kept up by a hooting Wl, and cabins on northern side are 1, 3 and 5 (see introduction and

inn) so his cabin was either 3 or 5, and one kept up by the hooting •Hi was either 1 or 2. The person kept up by croaking frogs was •Wcwhere west of the peach cabin, so not 5 or 6 and was exactly • li t the number of the cabin of the person kept awake by yowling

so these two cabins had to be even numbered cabins—thus, the W i n of the person kept up by croaking frogs was either 2 or 4, and B o n e of the person kept awake by yowling cats was either 1 or 2, so M l n s I and 2 were the cabins of individuals kept awake by a hooting 1*1 and yowling cats, in some order, so the person kept up by the ffltiikinj: frogs couldn't have cabin 2; he or she had cabin 4 and the one

•npi up by the yowling cats had cabin 2, which would leave the person tli i uhin I as the person kept awake by the hooting owl. The one kept

! • by the chirping crickets had a middle cabin, so could only have Hill cabin 3, which leaves the person kept awake by a barking dog, and Blr one kept awake by the buzzing fly in cabins 5 and 6, in some order. Wbt person in cabin 3 (chirping crickets) was between Mrs White's l lhln and the lemon-yellow one, so Mrs White's cabin was on the Hjtflh side, as was Col. Mustard's cabin. Mrs Peacock's cabin was next K |he one of the person kept awake by the buzzing fly (from above) B|ul directly across from Mr Green's cabin; so both Mrs Peacock Mid Mr Green had middle cabins, 3 and 4, in some order; thus, ^ nl Mustard and Mrs White (north side) had 1 and 5, in some order. Tins leaves Prof. Plum and Miss Scarlet with cabins 2 and 6, on the fimihcrn side, in some order.

I he lime-green cabin was on the same side as, and east of


Prof. Plum's cabin, so Prof. Plum had cabin 2 and Miss Scarlet ha! cabin 6 (see diagram). The sky-blue cabin is on the southern side, at is the lime-green one, and the lemon-yellow cabin is on the northern side. The person kept up by croaking frogs (cabin 4) had a cabin somewhere west of the peach cabin so either cabin 5 or 6 was peach The pink cabin has a smaller number than Mrs Peacock's (either 3 or 4) and larger than the lavender cabin. Since either Mrs Peacock or Mr Green had cabin 3 (north side) and Col. Mustard had a cabin on the northern side with a number more than twice that of the person kept up by the hooting owl (cabin 1), then Col. Mustard had cabin 5 and Mrs White had cabin 1; thus, Col. Mustard's cabin was lemon yellow. The peach cabin (somewhere west of the person kept up hy croaking frogs) had to have been Miss Scarlet's cabin, cabin 6. Sine* the peach cabin is cabin 6 (south side) and the lime-green one and J sky-blue one is also on the southern side (from above), then cabins I and 3 are pink and lavender, in some order. The pink cabin has a larger number than the lavender one. so the pink cabin is cabin a (crickets) and the lavender one is cabin 1. From clue 6, then, we set-Mrs Peacock can't have either the pink or lavender cabin, so the only possibility is she has cabin 4 and it is Mr Green who had cabin i j Since Mrs Peacock was next to the person kept up by the buzzing fly, then it was Miss Scarlet in 6 who was kept up by the buzzing lly, and by elimination, Col. Mustard in cabin 5 was kept awake by barking dogs. Prof. Plum didn't have the lime-green cabin, so he had J the sky-blue one, and by elimination, Mrs Peacock in 4 had the lime J green cabin.

In summary: (cabin number, guest, cabin's colour, sound) 1, Mrs White, lavender, hooting owl 2, Prof. Plum, sky blue, yowling cats 3, Mr Green, pink, chirping crickets 4, Mrs Peacock, lime green, croaking frogs 5, Col. Mustard, lemon yellow, barking dog 6, Miss Scarlet, peach, buzzing fly


ClUP Quilt

H i p arc two of each applique and positions are A through P. The H j i candlestick squares were in positions A and K; square C matches Ijtyiuc I, and square F matches square M. The two squares depicting

| i l Im 111 < are in corner positions, but not square A or M, so the squares KjUl the knife are in squares D and P. Three squares in a row, left to H n i , depict the revolver, rope, and green token are now in top row ^Whllesiick in A and knife in D) or third row (candlestick in K), so ^ • I n row 2 (E, F, G, H) or row 4 (M, N, O, P). One square with the H|m'h is to the immediate left of a square with the lead pipe, and the •ttU'i square with the wrench is to the immediate left of a square with

(Vvolver, and the two squares with the wrench touch at the corners, Hl l ie squares with the wrench are either in B and E or in squares I H d N (can't be L and O as no square to right of L). Whether one of • Squares with the wrench is in B or I, it is to the left of squares C

ftyl I which are matching (clue 1), so C and J are either the lead pipe revolver. However, since one of the squares depicting the revolver

H p Ihe immediate left of the one depicting the rope, which is to the H im-dialc left of the one depicting the green token, then the revolver •plild not be in either C or J (J in row 3 to the immediate left of the Hitllesiick); thus, squares C and J are the ones with the lead pipe fy|i|<lli|u6s.

•One of the squares with word 'Clue' is directly above the other W illi word 'Clue', so they can only be in squares E and I or in

Hpimes I I and L (see diagram), but since one of the squares with the Ken. Ii is in either E (and B) or I (and N), then the two with word Blue are in H and L. One of the squares with the knife (D and P) is in H f u w directly below a row which has a green token and since knife

|> is m top row, then it is the square with the knife in the bottom • lo which this refers, and so there is a green token in row 3, and it HI IK- in position I. Therefore, the two squares with the appliqud of Wrench can only be in B and E, and since C depicts the lead pipe, Ii F must depict the revolver, so square M also depicts the revolver.


Since one of the revolvers in F with 'Clue' in H is in row 2, then the three in a row in clue 5 are revolver in M, rope in N, and green token in O. By elimination, square G must depict the second of the squares with the rope.

In summary: (matching squares, symbol) A and K, candlestick B and E, wrench C and J, lead pipe D and P, knife F and M, revolver G and N, rope H and L, 'Clue' I and O, green token

Judy's Logic II

Thinking of You

Mr. Bluebell and each of the other five ClueJo characters •opened to receive a nice Thinking of You card from a different hlciul. Each received their card from one of the Clue suspects Who happened to be travelling somewhere in Europe at that • l ie , but was not going to be able to make it to England to visit •person. No two received the same type of card—each of which lind i different plant or flower as its theme. From the clues p low, determine the recipient and sender of each card, the M n i e of each card, and the country from which each card Htyiis postmarked (one is Italy).

| No Clue suspect sent a card which had a plant or flower on it that was the colour which was associated with his or In r name. l ive of the six recipients are Inspector Black (who didn't

I get a card from Mrs White), the person who received a card I'rom Prof. Plum, the person who received a card from Col. Mustard (who sent the card with red roses as its theme), the

I woman who received a card from Norway, and the person \s ho received a card postmarked from Greece.

Jl Sir Sable, who received a Thinking ofYou card with yellow


irises as the theme, isn't the person who received a card postmarked from Austria.

4. Miss Lavender, who didn't receive a card from Mr Grecri. received a card with lovely purple larkspur on it (which wasn't postmarked from Norway). Lady Melon received the card with exquisite white chrysanthemums.

5. The person who sent a card from Greece, who wasn't Mrs Peacock, picked out a card which featured feathery green ferns as its theme, while the card postmarked front Spain was received by Dr Gray.

6. The card with blue violets, which was postmarked front France, wasn't sent by Mr Green.

Table 7

RE CIPIENT CARD THEME COUNTRY

DR

GR

AY

IN

SPEC

TOR

BLA

CK

6 I >•

Q < J M

ISS

LAV

END

ER

MR

S B

LU

EB

EL

L SI

R S

AB

LE

BLU

E V

IOLE

TS

GR

EEN

FE

RN

S PU

RPL

E L

AR

KSP

UR

R

ED R

OSE

S W

HIT

E M

UM

S Y

ELLO

W I

RIS

ES

AU

STR

IA

FRA

NC

E G

RE

EC

E IT

ALY

N

OR

WA

Y

SPA

IN

SEN

DE

R COL. MUSTARD

SEN

DE

R

MR GREEN

SEN

DE

R

MISS SCARLET

SEN

DE

R

MRS PEACOCK

SEN

DE

R

MRS WHITE SEN

DE

R

PROF PLUM

CO

UN

TR

Y AUSTRIA

CO

UN

TR

Y

FRANCE

CO

UN

TR

Y

GREECE

CO

UN

TR

Y

ITALY

CO

UN

TR

Y

NORWAY CO

UN

TR

Y

SPAIN

AR

D T

HE

ME

BLUE VIOLETS

AR

D T

HE

ME

GREEN FERNS

AR

D T

HE

ME

PURPLE LARKSPUR

AR

D T

HE

ME

RED ROSES

AR

D T

HE

ME

WHITE MUMS U | YELLOW IRISES


Nont-a-Bus

ftnsl summer, Everett and his extended family had all planned • take their vacations around the same time. Camping was in B | c r , even for Everett's maternal and paternal grandparents, and •llhci than take several vehicles, they decided to use Rent-a-| t is and all travel together. Besides a lot of storage for all the lumping gear and food, and shelter in case of bad weather, the | I IS also had a men's and women's bathroom! On the way up Itnl Imck from their camping spot, each person sat in the same •.rat.

A day after returning from their vacation, Everett got a call (join a Rent-a-Bus employee saying that they had several items In the former to pick up which had been found aboard the bus— |n address book, a comb, a crossword puzzle magazine, an

Bye class case, a fashion magazine, keys on a ring, a lighter, a fcovcl, a monogrammed pen, and sunglasses. Each item had been Bund on a seat on the bus, inadvertently left behind by the person •fho had occupied that seat.

From the clues provided, determine the seat (lettered A through BB in the diagram) in which each person sat during (he trip, and the items left behind by ten of the twenty-one Occupants.

Bple: The members of the vacation group included Everett's Ba t e rna l grandparents—Marie and Emer; his paternal •mndparents—Sarah and Fred; his parents—Wilma and Calvin; Ills wife Sharon and their offspring—Megan and Shane; his

BUler Donna, Donna's husband Jerry, and their two offspring— Julie and Scott; his brother Norman, Norman's wife Sue, and llien two offspring—Seth and Sam; his brother John, John's wife I .on; and Sharon's nephew Eric, who drove the bus to and from

nhc vacation spot.


1. No married couple sat on the same side of the aisle. In only two instances did a person sit in a seat designated by a letiei which was the same as the first initial as his or her first name,

2. There were exactly two window seats and two aisle seats on each side of the aisle which remained vacant during the trip to and back from the vacation spot.

3. In the seats behind the driver, one of the three who occupied aisle seats, was the woman who forgot her comb (who sat al least three rows up from the woman who had left her address book on the bus).

4. Sue (who left behind a fashion magazine) sat to the immediate right of Donna, who sat directly behind Shane. John sat in a seat which was as far as possible, diagonally, from his grandmother, Marie.

5. Calvin sat to the immediate left of Everett and immediately behind Lori (who left behind a crossword puzzle magazine) The person who left behind the eyeglass case sat in the seal to the immediate left of Megan.

6. Seth, who had a window seat, sat next to an empty seat. Setli sat directly in front of Sharon, who sat to the immediate left of Wilma.

7. Sarah sat in seat P. Sam sat in seat B. The person who left behind the lighter sat in seat N. The person who left hei sunglasses on her seat sat in seat S. Seats A and H remained empty during the trip.

8. Jerry sat to the immediate right of Scott, who sat directly behind Fred, who in turn sat directly across the aisle from Julie (who didn't have a window seat).

9. The husband of the woman who sat in seat Y sat in seat F The wife of the man who sat in seat J sat in seat P. The wife of the man who occupied seat O had left behind a novel she was reading during the trip.

J u d y ' s L o g i c I I 71

III I xactly four people who left items on the bus had window seats, two on each side of the aisle. The occupant of seat I) left behind a monogrammed pen. One of Norman's sons left behind a set of keys.

- — A B C D E F G

Wom

en's

bath

room

N M L K J I H Wom

en's

bath

room

Aisle

Driv

er O P Q R S T U

Men

's ba

thro

om

Driv

er

BB AA Z Y X W V Men

's ba

thro

om

Figure 9

Iverett Goes Underground

Just this past March, on each Saturday of the month (2nd, 9th, I (.ill. 23rd and 30th), Everett and some of his family members Went spelunking, and each time they explored a different cave. Rheir guide described it as mountain climbing in reverse. They would use ropes and pylons and would be going down instead nl up. He urged them to descend in pairs and he'd be around in •KC of trouble.

Well, all went pretty well for the beginning spelunkers, and by the end of the month, they felt they could be called experienced. Ill fact, the only thing to spoil each weekend was that Everett had N I'ear of the strange and unusual, and while deep inside the depths nl the earth, he had occasion to see a few creatures which fitted that description. One rather strange type of animal was seen at


each location, and as Everett explained to his old friends, 'Nah, I wasn't scared. They just gave me the willies!' His family, of course, knew better!

From the following set of clues, determine the cave explored on each Saturday during March, Everett's partner each time, and what amazing and strange creature he saw on each occasion

1. Some very small, extremely rare white bats drove Evereti to take cover on the exploration they undertook on either 9 March or on the day that Everett partnered with his sister in-law, Shirley.

2. On the day that Everett was scared half out of his wits by a gigantic daddy-long-legs, his partner during the descent into the cave was Eric; this was sometime later in the month than the day that Everett took a foot-long earthworm to be a somewhat short, but venomous snake!

3. Everett partnered with his wife, Sharon, sometime later in the month than when they visited Hayle's Hole.

4. 23 March was either the day they visited Drake's Descent or it was the day that the white bats where spotted.

5. From first to last, but not necessarily on consecutive weekends, the family explored Colossal Caves, the cave where Shirley partnered with Everett, and the cave where a huge beetle with glowing antennae nearly gave Everett a heart attack.

6. It was either 9 March or it was the day that Everett almost fell in a shallow underground lake after nearly stepping on a harmless translucent scorpion that he partnered with his son, Shane.

7. Everett's daughter, Megan, and his nephew, Eric, were his partners when they visited Crystal Caves. Eric also partnered Everett the weekend immediately following the one when the white bats were seen.


The visit to Castle Caverns wasn't on 2 March, although it was some time earlier in the month that the cave they explored when Everett was partnered with Megan (which wasn't at Drake's Descent).

Table 8

TE DESTINATION PARTNER CREATURE

rse Play

ring one weekend in March, the Kasers stayed at the (nmding Resort in Echo Valley. One of the activities available

: h day is horseback riding, and the Kasers love to ride horses, i each day, Monday through Friday, one of the five adults— erctl, his wife Sharon, his brother Norman, his sister Donna, Donna's husband Jerry—went out with two of the young net s —Everett and Sharon's daughter Megan and son Shane, mua and Jerry's daughter Julie and son Scott, and Norman's a Scth.

I hey would have loved to ride as a family, but unfortunately, le were other guests at the ranch and so it was that on each V, only three horses were available. To decide which two of

five young people would ride, several straws were cut to


various sizes, and the two who got the longest and shortci(l straw would ride that day. How they decided on which adull would go is unknown and probably just as well!

Determine from the clues here, which adult and two young people rode each day, as well as the young person who drew! the longest straw and the one who drew the shortest straw.

1. Each young person ended up getting to ride on two differed I days of the week. On one day, he or she drew the long straw and on another day, he or she drew the shortest straw.

2. Megan drew the shortest straw the day immediately proceeding the day that Jerry rode (which was the day that Megan drew the longest straw).

3. Shane drew the longest straw the day immediately befora I the day Norman went out to ride.

4. On Tuesday, Julie drew the longest straw. The day Julie dre\l I the shortest straw was the day that Seth drew the longest straw, I

5. Scott drew the shortest straw the day that Everett rode. Don nil rode on Thursday.

6. Seth drew the shortest straw on Friday, which was not IliM day that Megan drew the longest straw.

Table 9 DAY ADULT LONGEST STRAW

DRAWN BY SHORTEST STRAWl

DRAWN BY


Walking the Appalachian Trail

M i l long ago, Everett Kaser and his family had a chance to fflilk part of the Appalachian Trail (well, actually only 20 miles

H f ll), but they wanted to be prepared for anything, so besides ^ p i c r and some food and extra clothing in a knapsack that each Mirricd, each person was assigned one item to be sure and bring

—just in case! Also, each person brought his or her own wtiiuTa and each took a different number of pictures along the Hfyy (from 10 to 18). The family members included Everett, his Iwilc Sharon, son Shane, daughter Megan, Everett's sister Donna, U)iinna's husband Jerry, Donna and Jerry's son, Scott, Sharon's Hpicr , Shirley, and Shirley's son, Eric.

Hie nine family members looked a colourful bunch as they j j t l out walking, single file, in different coloured shirts—blue, • i r c n , lavender, orange, pink, red, tan, white, and yellow. Find r Mil Mom the clues given under, the order that each member •Hurled out walking the trail, the colour of shirt each wore, the

IIII |H nlant item each brought along, and the number of snapshots I Win Ii of them took.

I The four immediate members of Everett's family are Megan, I lie individual who took 15 snapshots, the individual who was next to last in line as they began their hike, and the individual who wore the blue shirt.

I, If n y, who brought along a compass, took a total of 11 pictures. Scott took a total of 13 pictures.

RI Sharon took one more snapshot than Shane, who took one ! more snapshot than the person who brought along a pocket

knife, who took one more snapshot than Everett, who took one more snapshot than the person who wore the lavender shirt (who was number seven in the group when they started out).


4. The person who took 12 pictures wasn't the male member <>1 the family who wore the yellow shirt. Neither Shirley (who started out immediately behind the person who brought the toilet paper) nor Sharon is the person 'who brought along » flashlight.

5. Shane was first in line when they started out, while the person who carried the pocket knife was fourth. The sixth person in line was a male. The fifth person in line wasn't the female in the white shirt (who also wasn't Sharon).

6. As the group set out, five family members walking in single file, consecutively from the first to the last, were Eric (win carried the first-aid kit), the female in the pink shirt (wlui wasn't Megan), Scott, the person in the tan shirt who carried

Table 10 ORDER FAMILY

MEMBER SHIRT

COLOUR ITEM

BROUGHT SNAPS TAKEN


the bug spray, and the person who had brought along a battery-operated radio.

I I he third person in line (who wore a green shirt) took more photographs than his mother, who brought along some zip-lock bags. The person in orange shirt, who brought matches, look more pictures than the second person in line, but fewer than the person in the red shirt.

Skydiving

Ijist weekend, the Kasers went skydiving. Everett's sister Donna ami her husband, Jerry, accompanied Everett, along with his wife Sharon, son Shane, and daughter Megan. The group picked out Iheir parachutes from a colourful selection at the skydiving lei niinal—blue, green, orange, purple, red, or yellow, with each luinily member selecting a different colour. It was a beautiful day. and all went quite well... er... until they landed on earth, Ihnt is. Each person landed in a completely different location lioni where he or she had intended to land—one individual landed In a pig's sty! From the clues given here, determine the order In which each person left the plane (the jumpers alternated by gender), the location of each person's landing, and the colour 01 each person's parachute.

I, Sharon jumped immediately after the person who landed in a fish pond and immediately before the person with the green parachute.

2 I he three men were Jerry, the person who wore the red parachute, and the person who jumped immediately after the individual who landed in a haystack. It was not the person who had the blue parachute who was the last of the jumpers.


4. Donna jumped some time after the person with the orange parachute, who landed in a strawberry patch.

5. It was not the person who had the green parachute who landed in a cow pasture and narrowly escaped the bull grazing there'

6. The family member who had the unfortunate luck to get tangled in a large oak tree in a farmer's front yard had a somewhat uncomfortable wait until someone arrived who could get her down.

7. Three jumpers in consecutive order, from first to last, were the person with the purple parachute, one of the Kaser's siblings, and Everett.

Table 11 ORDER FAMILY

MEMBER LANDING LOCATION

PARACHUTE COLOUR

The Dunk Tank

When Stefano's niece, Elizabeth, approached him aboul participating in her school's upcoming fundraiser, he was rather flattered. He found that during the afternoon in question, he would be in the Dunk Tank, and as the weatherman had promised a


cither exceptionally hot day, Stefano hoped to get dunked as illicit as possible!

The cost was one ticket per throw of the ball, unless of course you hit the mark. Then one got a free throw. During the two hours that Stefano spent in the Dunk Tank, there were five lllllerent children who ended up throwing the ball several times lot the cost of one ticket (2 ,3 ,4 , 5 or 6 times). Each child was in i different grade—second through sixth.

To encourage the children, Stefano would throw taunts at litem. However, in each of the five cases mentioned above, the child (one of whom is Kevin) was laughing so hard, he or she finally missed the mark and failed to dunk Stefano. From the liven clues, determine the name and grade level of each child who got more than their money's worth out of each ticket, the Dumber of times that each ended up throwing the ball, and the final taunt that finally ended their streak of luck!

I, Stefano's taunt of 'Hey, shortie, your owner know you're off the leash?' caused Emily to miss the paddle by several inches. The second-grader is neither Thomas nor the child whose final taunt was 'Yo, kid, those freckles or have you got the measles?' Carrie is in the fourth grade.

I The fifth-grader ended up throwing the ball the most times before his or her streak came to a screeching halt, mainly due to a case of the hiccups from laughing so hard.

I It was the sixth-grader whose throwing streak came to an end after only two throws of the ball with the taunt 'Whoa there, kiddo, only earthlings are allowed to play!'

J, Emily is in one grade higher (and got in exactly two throws more) than the child whose final taunt from Stefano was Wow, those ears flap in the wind much, kid?'


6. Uriah, who isn't the second-grader, doubled over in laughtei with the taunt 'Say, kid, I've seen better arms than yours on a rocking chair!' and didn't even bother trying to throw again

Table 12 CHILD'S NAME

GRADE NUMBER OF THROWS

STEFANO'S TAUNTS

Berry Delicious!

Recently, Stefano celebrated the tenth anniversary of his promotion to maitre d' at Gordino's Restaurant, and was given the week oil, On each of the five days of that week, Monday through Friday, Stefano was invited over for supper at a different relative's home where, in honour of the occasion, he enjoyed one of his favourite treats—berries! Stefano loves berries of all types, and at each home he was served a different type of berry, including blackberry, made in a different type of food. Stefano showed up at the home of each relative with a different special gift—a book, chocolates, flowers, perfume, or potted plant.

Can you determine from the clues given under, the day of the week he visited each relative—aunt, grandmother, mother, niece, or sister, the gift he presented each one, and the treat Stefano enjoyed at each home?


I The day Stefano presented his hostess with a lovely potted plant was exactly two days before he was invited to his mother's home, which was the day before he enjoyed a treat made with raspberries.

I. It wasn't Stefano's aunt who received the book of poetry. 3. Stefano had pancakes the day before he visited the relative to

whom he gifted flowers, which was the day before he visited his sister.

I Neither muffins nor pie were the special treat served the day that Stefano presented his hostess with a potted plant.

.V Neither cranberries nor blueberries was the main ingredient in either the pie nor the muffins Stefano enjoyed that week,

Table 13

RELATIVE GIFT GIVEN BERRIES DESSERT

£ 3

1 G

RA

ND

MO

TH

ER

I

1 M

OTH

ER

[ NIE

CE

1 SIS

TER

1

BO

OK

1 C

HO

CO

LATE

S I F

LOW

ERS

1 PE

RFU

ME

1 PO

TTED

PL

AN

T I

1 B

LAC

KB

ERR

Y

1 BLU

EBER

RY

1 C

RA

NB

ER

RY

1

RA

SPB

ERR

Y

1 ST

RA

WB

ERR

Y

1 CO

BB

LE

R

1 MU

FFIN

S 1 P

AN

CA

KES

U £ |

PUD

DIN

G

MONDAY ft* TUESDAY . "J WEDNESDAY . "J

THURSDAY FRIDAY COBBLER MUFFINS

Vi PANCAKES PIE PUDDING BLACKBERRY

JERR

IES

BLUEBERRY

JERR

IES

CRANBERRY

JERR

IES

RASPBERRY STRAWBERRY

sr. BOOK HJ £ O

CHOCOLATES HJ £ O FLOWERS

t PERFUME o POTTED PLANT


nor did Stefano enjoy cranberry cobbler at any home that week.

6. On three consecutive days, Stefano ate some delicious cobbler, visited the home of his niece, and enjoyed a treat made with blueberries (which wasn't at the home where he presented his hostess with a box of chocolates).

7. The day he was invited to his grandmother's home wasn't the day he enjoyed a wonderful berry pie. It was not at his grandmother's home that he had pancakes, nor was it at his sister's home that he had the pie.

8. The five days that Stefan had off that week were Thursday, the day he was invited to his grandmother's, the day he enjoyed a treat made with strawberries, the day he enjoyed pudding made with one of his favourite berries, and the day he presented his hostess with a box of chocolates.

Unique Friends

Last year, Stefano vacationed in New England and managed to visit five old High School buddies. Each friend had a very unusual first name—for having been born in the twentieth century that is, and so each went by a different nickname. Not only did each have a rather unusual first name (one is Aloysius), but each also lives in a rather different type of dwelling (one lives in a converted boxcar). From the given clues, determine each of Stefano's friends' first name, nickname, type of dwelling, and the New England state in which each one lives.

1. The five friends are the man nicknamed Jinx, the man who lives in Delaware, the man who lives in New Hampshire, the man who calls a converted windmill home, and the man who calls a converted old barn his home.


2. Neither Cathimor nor Dedrick is the man nicknamed Marbles or the man who resides in New Jersey.

V, Neither Baldric nor Dedrick (neither of whom live in Vermont) are the friend known by the nickname Egghead (who does not live in the converted windmill).

•1. Neither the Vermont resident nor the Connecticut resident are known by the nicknames Whistler or Jinx, and none of these four gentlemen is the friend Stefano visited whose home is a tree house.

5. Ebenezer's nickname is neither Torpedo nor Egghead, and none of these three men live in the tree house or the converted cave.

Table 14

NICKNAME ABODE STATE

[EG

GH

EA

D

| JIN

X

[MA

RB

LE

S

[TO

RP

ED

O

[WH

IST

LE

R

| B

OX

CA

R

| C

AV

E

| O

LD

BA

RN

| T

RE

E H

OU

SE

| W

IND

MIL

L

| C

ON

NE

CT

ICU

T

| D

EL

AW

AR

E

| N

EW

HA

MP

SH

IRE

| N

EW

JE

RS

EY

VE

RM

ON

T

A L O Y S I U S

B A L D R I C

< z C A T H I M O R < z D E D R I C K

E B E N E Z E R

C O N N E C T I C U T

g D E L A W A R E

g N E W H A M P S H I R E

b N E W J E R S E Y

V E R M O N T

B O X C A R

U J C A V E c o O L D B A R N CQ < T R E E H O U S E

W I N D M I L L


6. When visiting Aloysius and Dedrick (neither of whom is nicknamed Jinx), Stefano travelled to Delaware in one instance, and stayed in a comfortably furnished cave in the other case.

7. Stefano played football in high school with both Cathimor and the man who resides in Connecticut.

Strange Dreams

Stefano has worked at Gordino's Italian Restaurant for many years, and has been proud of the fact that it has always received a five-star rating from the various restaurant critics around the country, which is the highest rating a restaurant can receive. In the past year, five new restaurants have opened in his city, each serving a different type of food—Chinese, French, German, Hungarian, or Swedish. Being the curious type, Stefano decided to dine out at each one to see how their service and quality of food compared to Gordino's. He dined out at a different restaurant during one week, Monday through Friday, and on Saturday gave the owner of Gordino's his opinion of each one.

Stefano gave each a different rating (from a one-star to a five-star restaurant). At each restaurant, he paid a different amount for each meal ($8.00, $8.50, $9.00, $ 10.00 and $ 12.00), which he found quite reasonable. However, after dining out at each restaurant, that same night he had a very strange dream Each night he had a different dream, and they were so weird he decided against telling the owner of Gordino's about them!

Determine from the following clues, the evening he dined at each restaurant, the price he paid for each meal, the rating-he gave each restaurant, and the strange dream he experienced each night after retiring.

1. Stefano's evening meal cost more the night he ate at the Hungarian restaurant than it did at both the restaurant he


ate at on Thursday evening, and the night he dreamt that he was seated in a restaurant where he and everyone else in the restaurant had clothes on that were way too small for them. In his dream, he could even hear ripping sounds as the back end of men's pants split out as they sat.

K . The meal Stefano ate at the Swedish restaurant was later in the week than the meal he paid $9.00 for, which was later in the week than the evening he dreamt he was stuck to his chair with super glue and couldn't get his wallet out to pay for the meal or leave a tip (which was the night he rated the restaurant he dined at as a three-star restaurant).

• 3 . Stefano paid more for the meal he had at the French restaurant than he did at both the Swedish restaurant and the restaurant where later in the evening he dreamt he had ordered piles of boxes of Cracker Jacks, looking for a prize in each one which would help him pay for his meal (which was earlier in the week than when he dined at the French restaurant).

[ 4. He rated the German restaurant one star lower than he did the restaurant he dined at the night he dreamt that everyone's clothes were too small. Stefano paid more for his meal at the German restaurant than he did at the one where later he dreamt everyone's clothes were too small, but less than he paid on Tuesday night. The restaurant he dined at on Tuesday night he gave a four-star rating.

W3. The five restaurants are the Chinese restaurant, the restaurant where he dreamt of the Cracker Jacks, the one he ate at on Thursday night, the restaurant where he paid $8.00 for his meal, and the restaurant he gave a five-star rating.

R 6. Stefano paid more for his meal on Monday night than he did the night he ate at the French restaurant, or the night he dreamed he had shrunk to only 2 inches tall and was swimming in a bowl of chicken noodle soup (which was on Thursday night).


7. The restaurant he dined at the night he dreamt that no one in the place had any silverware and everyone was looking around to see how others were going about eating their meal was the night he gave a rating that was exactly three-stars more than the restaurant he dined at on Wednesday night.

Table 15

DAY TYPE OF RESTAURANT

PRICE PAID RATING DREAM

Aunt Agatha's Haunted House

Stefano had not actually seen his Aunt Agatha since he was II child, but he had heard his family talk of his eccentric aunl who claimed to live in a haunted house. When Aunt Agatha invited Stefano to visit one week during the holidays, he jumped at the chance. Two things he discovered—one is that his aunt was indeed eccentric, and the other—she did indeed live in ii haunted house. On the five weekdays he was there (Monday through Friday), at exactly five minutes before midnight, a ghostly apparition appeared, each in a different room of the house Each sighting was preceded by a different sound (one sound was whistling).

His aunt claimed they were members of the staff of the


•previous owner who had mysteriously disappeared and were awaiting his return. However, Stefano neither quaked nor

ll quivered, and it was many years later that he found, because of t his intrepid acceptance of the ghosts, his Aunt Agatha had left

him her entire (and very substantial) fortune! From the following • clues, determine the day he saw each ghost, the room where he • NUW each of them, and the sound that preceded each sighting.

1 1 . Wednesday night was the night a ghostly appearance was preceded by a mournful moaning sound.

1 2 . The ghost Stefano saw on Thursday evening (which was not the ghost of the housekeeper) appeared in his Aunt Agatha's old-fashioned kitchen.

• 3. Neither the ghost of the parlour maid nor the ghost of the housekeeper is the one Stefano saw on Friday evening.

1 4 . The sounds of hiccupping preceded the appearance of the upstairs maid, which wasn't in the parlour.

13. On Tuesday night, as Stefano wandered through the vast rooms of his aunt's huge old house, he heard crying and immediately afterwards he sighted the ghost of the cook.

Table 16 DAY APPARITION

SEEN ROOM SOUND


6. Near the end of the week, the ghost of the butler appeared in the guest bedroom as Stefano made preparations to retire

7. A tap-tap-tapping sound, which didn't herald the appearance of either the butler or the housekeeper, occurred on the night just after the night Stefano saw a ghost in the dining room

Solutions

Thinking of You

Five of the recipients were Inspector Black, one who received a can! from Prof. Plum, one who received a card from Col. Mustard, one who received a card from Norway, and one who received a card from Greece The person who received the card from Norway was a female, and Col. Mustard sent the card with red roses, and the person who seni the card from Greece sent the card featuring green ferns. Miss Lavender got the card with purple larkspur, so it wasn't sent by Prof. Plum, nor was her card postmarked from Norway, so she isn't one of the live people mentioned in clue 2—she is the sixth recipient. Sir Sable received the card with yellow irises (clue 3), so he isn't the person whi> got a card from Norway (sent to a female), so he must be the one who got a card from Prof. Plum. Therefore, Lady Melon, who received u card featuring white chrysanthemums, is the female who received u card postmarked Norway, and by elimination, Inspector Black received the card with blue violets. The card with the blue violets (received by Inspector Black) was postmarked from France, but wasn't from Mr Green, nor was the card with the green ferns nor the card received by Miss Lavender (purple larkspur); thus, Mr Green can only be the one who send Lady Melon (white chrysanthemums, Norway) a can I The card from Spain was received by Dr Gray, so he wasn't the on* who received the card with the green ferns (Greece); thus, Dr Gray is the one who got a card from Col. Mustard (red roses), and by elimination, Mrs Bluebell must have gotten the card with the green


lems. Sir Sable (from Prof. Plum, yellow irises) didn't get the card postmarked from Austria so his card was postmarked from Italy (introduction), and it was Miss Lavender whose card was postmarked Austria. Mrs Peacock (associated with colour blue) didn't send the Card with the blue violets to Inspector Black, nor did he receive his Card from Mrs White; Inspector Black's card was received from Miss Scarlet. Mrs Peacock also didn't send a card from Greece so she didn't send it to Mrs Bluebell; the card from Mrs Peacock went to Miss Lavender, and by elimination, Mrs White is the one who sent the Card with the green ferns, postmarked Greece, to Mrs Bluebell.

In summary: (recipient, sender, theme, postmarked from) Dr Gray, Col. Mustard, red roses, Spain Insp. Black, Miss Scarlet, blue violets, France Lady Melon, Mr Green, white chrysanthemums, Norway Miss Lavender, Mrs Peacock, purple larkspur, Austria Mrs Bluebell, Mrs White, green ferns, Greece Sir Sable, Prof. Plum, yellow irises, Italy

Rent-a-Bus

Items left behind are address book, comb, crossword puzzle magazine, eyeglass case, fashion magazine, keys on a ring, lighter, novel, monogrammed pen, and sunglasses. Seats are designated A through III! (diagram). A list of the passengers is given in the introduction. Nai ah sat in seat P; Sam in seat B, person who left the lighter in seat N, the person who left sunglasses in seat S, and A and H remained empty neats. John sat as far as possible, diagonally, from Marie and since no One sat in seat A, then John and Marie sat in seats G and BB, in some order. The wife of the man who sat in seat J sat in seat P, and since we determined Sarah sat in seat P, then it was Fred who sat in seat J. Jerry sat to the immediate right of Scott, who sat directly behind Fred, who sat directly across Julie, so Jerry sat in seat F, Scott in seat I, Megan in seat E, and since Julie didn't have a window seat (clue 8)

90 100 M i n d s t r e t c h

then Julie sat in seat S, and is the one who left the sunglasses. The person who left behind the eyeglass case sat in the seat to the immediate left of Megan (seat E), so Fred in J left behind the eyeglass case. Since Jerry sat in seat F, then it was Donna who sat in seat Y Sue sat to the immediate right of Donna, who sat directly behind Shane. and since we just determined Donna had seat Y, then Sue was in seal R and Shane in seat Z. Sue, in seat R, left behind a fashion magazine Seth sat in front of Sharon, who sat to the immediate left of Wilma, and since Seth had a window seat, then for Sharon to sit to the immediate left of Wilma, they had to sit on the side of the bus behind the bus driver, who was Eric. Seth sat beside an empty seat, so didn't have seat BB (John or Marie), seat AA (Sarah in P); or X (Julie in S left sunglasses); so the only possibility is that Shane sat in W, and T was empty, Sharon sat in V, and Wilma in U.

A man sat in seat O, and since two seats on each side of the aisle were empty, then on the side behind driver, Q was empty during tin-trip. And since either John or Marie sat in seat BB, then seats AA and X behind the driver were empty seats. Calvin sat to the immediate left of Everett and immediately behind Lori, and since both Sharon (Everett's wife) and Wilma (Calvin's wife) sat on the side behind tin-driver (Sharon in V and Wilma in U), then Everett and Calvin sat in seats across the aisle from seats behind the driver. And since Lori is married to John, then John must be the one who had seat BB and Marie had seat G. Exactly two people who sat in seats that were designated with a letter that is the same as his or her first initial (clue 1). There are three people with the first initial J (Jerry, John, and Julie), but Fred had seat J, and Everett nor Emer were in seat E (Megan was in I i, Marie wasn't in M (she was in G), Wilma was in U and Shane in W. Donna wasn't in D (she was in Y), and none whose first initials were S were in S (Julie was in S); thus, the only possibilities are that Norman was in N and Lori in L, so Norman in N was one who left behind his lighter. Since Lori was in L, then Everett was in D and Calvin in K, and Lori is the one who left behind a crossword puzzle magazine By elimination, the only seat left occupied was O and had to be when-


Emer sat. On that side of the bus, opposite the driver, since the aisle seats N, L, K, J, and I were occupied, then M was an empty seat (along with H), and since window seats B, D, E, F, and G were occupied, then C was empty. Thus, at this point we have A empty, B is Sam, C empty, Everett in D, Megan in E, Jerry in F, Marie in G, H empty, Scott in I, Fred in J (left eyeglass case), Calvin in K, Lori in L, M empty, Norman in N (left lighter), Emer in O, Sarah in P, Q empty, Sue in R I left fashion magazine), Julie in S (left sunglasses), T empty, Wilma

1 in U, Sharon in V, Seth in W, X empty. Donna in Y, Shane in Z, AA empty, and John in BB. So we have six of the ten items found on the bus left to place. Everett in seat D is the one who left behind the tnonogrammed pen. The wife of the man in seat O left behind the novel, so that is Marie in G. One of Norman's sons left behind keys, .ind his sons are Seth and Sam. Since exactly two people who had window seats on each side of the aisle left behind items, and Everett in D left behind the pen and Marie in G left the novel, then Sam in B didn't leave behind an item; that would be Seth in seat W who left behind the keys. Since on each side of the aisle, two occupants by the window seats left items, and ten items were left, then the three who had aisle seats on each side left behind an item.

On the side opposite to driver's side, the three were Norman, Lori and Fred, and on the side behind the driver, we have Seth in window seat W leaving keys, and in aisle seats, Sue in R left behind the fashion magazine and Julie left behind the sunglasses. On the driver's side, the woman who forgot the comb, sat in an aisle seat at least three rows up liom woman who forgot the address book, so it could only be Sarah In I' who left behind the comb, and the other two on the aisle seats are Sue who forgot the fashion magazine and Julie who forgot the sunglasses (all from above), so the woman who forgot the address book had a window seat and must have been Sharon.

In summary: (seat, person, item left behind) A, empty B, Sam C, empty


D, Everett, monogrammed pen E, Megan F, Jerry G, Marie, novel H, empty I, Scott J, Fred, eyeglass case K, Calvin L, Lori, crossword puzzle magazine M, empty N, Norman, lighter O, Emer P, Sarah, comb Q, empty R, Sue, fashion magazine S, Julie, sunglasses T, empty

Everett Goes Underground

Eric partnered with EK the day E saw the gigantic daddy-long-legs EK's partner on the first outing was not Eric, Sharon, Shirley, or Megan, it was Shane, and that was the day he saw the translucent scorpion EK's partner on 9 March wasn't Shirley, Eric (after E saw the foot-long worm), or Megan (after Castle Caverns which weren't the firsi caves explored); it was Sharon, so it was Hayle's Hole that they explored as the first outing on 2 March. EK's partner the day he saw the huge beetle with the glowing antennae wasn't Sharon (right after Hayle's Hole) or Shirley; it was Megan. Megan (beetle with glowing antennae i wasn't EK's partner during the exploration of the Colossal Caves, Castle Caverns or Drakes Descent, so she was his partner during the week they visited the Crystal Caves. Megan (Crystal Caves, beetles) didn't partner with EK on 9 or 16 March (2 March, Shane, Hayle's Hole) oi on the 23rd, so she was his partner on 30 March, and that was the day


they visited the Crystal Caves and saw the beetle with glowing antennae. Therefore, Eric and Shirley were EK's partners on 16 and 23 March, in some order. Eric partnered with EK the weekend immediately following the one when they saw the white bats; so the white bats were not seen on 23 March; thus, that was the day they explored Drake's Descent (clue 4). On the visit to Drake's Descent on 23 March, EK didn't see the white bats or the foot-long worms, so it was the day that EK saw the gigantic daddy-long-legs, and Eric was EK's partner that day; therefore, the white bats were seen on the 16th, which leaves Sharon (9 March) as his partner the day he saw the foot-long worms. By elimination, Shirley was his partner on the 16th and the day they explored the Colossal Caves was on 9 March. By elimination, it was on the 16th that they explored Castle Caverns.

In summary: (date, place, partner, creature) 2nd, Hayle's Hole, Shane, translucent scorpion 9th, Colossal Caves, Sharon, foot-long worm 16th, Castle Caverns, Shirley, white bats 23rd, Drake's Descent, Eric, gigantic daddy-long-legs 30th, Crystal Caves, Megan, beetle with glowing antennae

Horse Play

Megan drew the shortest straw the day immediately before the day Jerry rode, which was the day Megan drew the longest straw, so the two days that Megan rode were consecutive days. She didn't draw the longest straw on Monday or Tuesday, so she didn't get the shortest one on Monday, and she didn't draw the longest straw on Friday, so didn't draw the shortest one on Thursday; and since Donna rode on Thursday, then Megan didn't draw the longest straw that day, so didn't get the shortest on Wednesday. Therefore, it was on Tuesday that Megan drew the shortest straw, which is also the day that Julie drew the longest straw, and on Wednesday she drew the longest straw, so Wednesday is (he day Jerry rode. Seth drew the shortest straw on Friday. Scott drew the shortest straw the day Everett rode, so it wasn't on a Friday,


Thursday, Wednesday, or Tuesday; thus, Everett rode on Monday and Scott drew the shortest straw on Monday. Julie drew the shortest straw on the day that Seth drew the longest one; so the only possibility left that we haven't determined with shortest, longest, or both is for Thursday. Thus, on Thursday when Donna rode, Julie drew the shortest straw and Seth drew the longest. Since the shortest straws were drawn by Scott on Monday, Megan on Tuesday, Julie on Thursday, and Seth on Friday (all from above), then Wednesday is the day that Shane drew the shortest straw. Shane drew the longest straw the day before Norman rode which wasn't Friday (Seth longest, Julie shortest), so Norman rode on Tuesday and Shane drew the longest straw on Monday. By elimination, it was on Friday that Sharon rode and Scott drew the longest straw.

In summary: (day, adult, longest straw drawn by, shortest straw drawn by)

Monday, Everett, Shane, Scott Tuesday, Norman, Julie, Megan Wednesday, Jerry, Megan, Shane Thursday, Donna, Seth, Julie Friday, Sharon, Scott, Seth

Walking the Appalachian Trail

Shane was first in the line, the person carrying the pocket knife was fourth, and the sixth person was a male, while the third person in the line wore a green shirt and was also a male and the seventh person in the line wore a lavender shirt. The third person in the line had a mother, so he was either Shane, Scott, or Eric (introduction), but not Shane (clue 5), so was Eric or Scott. As the group set out, five people, in consecutive order from first to last, were Eric (first-aid kit), female in pink shirt, Scott, person in the tan shirt who carried the bug spray, and the person who brought the battery operated radio; so since Shane-was first, and either Eric or Scott was third, and according to clue 6, Eric was one person ahead of Scott, then Eric was the third person in

J u d y ' s L o g i c I I 94

the line, who wore a green shirt and carried the first-aid kit, a female in pink shirt was fourth and carried the pocket knife, Scott was fifth, the person in the tan shirt who carried the bug spray was sixth and was a male, and the person who brought the radio was seventh in line.

Sharon took one more picture than Shane, who took one more picture than the person with the pocket knife (fourth in line, pink, female), who took more pictures than Everett, who took one more picture than the person in the lavender shirt who was seventh in the line; thus, Sharon either 18, 17, 16, 15, or 14, but not 14 as Shane would have taken 13. or 15, as Shane then took 14 and female in pink shirt with pocket knife took 13, or 16 as Shane then took 15, female in pink 14. and Everett 13: thus, Sharon took either 17 or 18 pictures, Shane took 16 or 17, female in pink took 15 or 16, Everett took 14 or 15, and person in the lavender shirt took 13 or 14. Four members of Everett's immediate family (Everett, Sharon, Megan and Shane) are Megan, the person who took 15 photos, the eighth person in line, and the person in the blue shirt. The eighth person wasn't Megan or Shane, so was either Sharon or Everett. Since there are five males, and Shane was first, Eric second, Scott third, and the person who was sixth (tan shirt and bug spray) was a male, but not Jerry (brought compass), then Everett could only be sixth in line, which leaves Sharon as eighth. Since Sharon took either 17 or 18 pictures, and Shane took either 16 or 17, then the person who took the 15 pictures was Everett, and since Sharon was eighth in line, then it was Shane who had on the blue shirt. Now that we know Everett took 15 pictures, then Sharon took 18, Shane 17, female in pink shirt with pocket knife took 16, and the person in the lavender shirt took 14. Thus, so far we have First—Shane, blue. 17; Second—unknown; Third—Eric, green, first-aid kit. Fourth— female in pink, pocket knife, 16; Fifth—Scott, 13; Everett, tan, bug spray, 15; Seventh—lavender, radio, 14; Eighth—Sharon, 18: and Ninth—unknown. Thus, we know who took from 13 to 18 pictures, leaving 12, 11, and 10 to be accounted for. Jerry took 11 pictures, and third person took more than his mother, so he could only have taken 12 and his mother, Shirley, took 10 and brought zip-lock bags.


Therefore, Jerry and Shirley were second and ninth in line, in some order, which leaves the female in fourth and person in seventh as Donna and Megan, in some order. The one in the pink shirt wasn't Megan so it was Donna, and Megan wore the lavender shirt. One female wore a white shirt, but not Sharon and since Donna wore pink and Megan lavender (from above), then it was Shirley who wore the white shirt. One male member wore a yellow shirt, and since Sharon took the most pictures, then she didn't wear the orange shirt, so she could only have worn the red shirt. Since Jerry brought the compass and Shirley brought the zip-lock bags, the person who wore the orange shirt who brought matches can only be Scott. Sharon didn't bring the flashlight and since we know Jerry brought the compass and Shirley zip-lock bags, then Sharon brought the toilet paper, and since she was eighth in line, then Shirley was ninth; thus, Shirley was ninth and Jerry was second and by elimination, Jerry wore the yellow shirt and Shane is the one who brought the flashlight.

In summary: (number in line, name, colour of shirt, item, number of suapshots)

First, Shane, blue, flashlight, 17 Second, Jerry, yellow, compass, 11 Third, Eric, green, first-aid kit, 12 Fourth, Donna, pink, pocket knife, 16 Fifth, Scott, orange, matches, 13 Sixth, Everett, tan, bug spray, 15 Seventh, Megan, lavender, radio, 14 Eighth, Sharon, red, toilet paper, 18 Ninth, Shirley, white, zip-lock bags, 10

Skydiving

There are three women—Donna, Megan and Sharon, and three men Everett, Jerry, and Shane and the jumpers alternated by gender (introduction). The person who had the orange parachute landed in a


strawberry patch. The first jumper wasn't Sharon, Donna Everett, and Megan (one of the Kaser siblings) and since just before Everett, has to be a female; thus, the first jumper was either Shane or Jerry. So, the first, third, and fifth jumpers were males, and the second, fourth, and sixth jumpers were females. Neither Sharon nor Megan was last, so the last jumper was Donna. Since Sharon jumped immediately before the person with the green parachute, then the one with the green parachute was a male, as was the one who had the red parachute, and the person who had the purple parachute (immediately before Megan); thus, the females had the blue, orange and yellow parachutes. Donna (sixth, last) didn't have the blue parachute or the orange one, so Donna's parachute was yellow. The females were the ones who landed in the haystack (jumped right after a male), the one with orange parachute who landed in a strawberry patch, and the one who landed in an oak tree, so the males were the ones who landed in the fish pond, cow pasture, and pig sty. Since Sharon jumped immediately after the male who landed in the fish pond and immediately before the male who had the green parachute which wasn't the one who landed in the cow pasture, then the one with the green parachute is the male who landed in the pig sty.

Everett didn't have the purple parachute, and since Sharon jumped immediately before the one with the green parachute, and Megan jumped immediately before Everett, then Everett didn't have the green parachute—Everett's parachute was red, so by elimination, Shane is the male who jumped immediately before the female who landed in a haystack. The male who had the purple parachute jumped immediately before Megan and the male who landed in a fish pond jumped immediately before Sharon; so the male who had the purple parachute didn't land in the fish pond; therefore, it was Everett (red) who landed in the fish pond. Thus, between clues 1 and 7, and since we know that Donna (yellow) jumped last, then the one with the purple parachute went first, Megan second, Everett third, Sharon fourth, and the male with the green parachute who landed in a pig sty, fifth. Thus, the first


jumper is the one who landed in the cow pasture. Everett didn't jump immediately after the female who landed in a haystack, nor did Jerry, so Shane is the one who jumped right after the female who landed in a haystack, who can't be Megan; and it can't be Donna, so Shane jumped right after Sharon; therefore, Sharon landed in a haystack, and Shane is the one with the green parachute who landed in a pig sty, and by elimination, Jerry was the first jumper (cow pasture, purple). Since Donna didn't have the orange parachute and land in a strawberry patch, then that was Megan, and by elimination, Sharon had the blue parachute and Donna landed in the oak tree.

In summary: (order of jump, name, place of landing, colour of parachute) First, Jerry, cow pasture, purple Second, Megan, strawberry patch, orange Third, Everett, fish pond, red Fourth, Sharon, haystack, blue Fifth, Shane, pig sty, green Sixth, Donna, oak tree, yellow

The Dunk Tank

The second-grader is not Tommy, Carrie, Emily, or Uriah, so it is Kevin (introduction). The taunt to Kevin was not 'Hey Shortie, your owner know you're off the leash?' (that was to Emily), 'Yo, kid, those freckles or have you got the measles?', 'Whoa there kiddo. only earthlings are allowed to play!' (sixth-grader), or 'Say kid, I've seen better arms than yours on a rocking chair!'; thus, the taunt to Kevin was 'Wow, those ears flap in the wind much, kid?'. Taunt to Emily was 'Hey Shortie, your owner know you're off the leash?'. Taunt to Uriah was 'Say kid, I've seen better arms than yours on a rocking chair!'. Carrie is in the fourth grade. The sixth-grader threw only two balls and got the taunt of 'Whoa there kiddo, only earthlings are allowed to play!', so, by elimination, Carrie in the fourth grade got the taunt 'Yo, kid, those freckles or have you got the measles?', and by elimination the sixth-


grader was Tommy. This leaves Emily and Uriah as the third and fifth graders. Emily is in one grade higher than the child whose taunt was 'Wow, those ears flap in the wind much, kid?' which was Kevin, so Emily is in the third grade, and Uriah in the fifth grade. Uriah threw the ball the most times (6 times). Emily got two throws more than Kevin, and since Uriah had 6 throws and Tommy only 2 (both from above), then Emily and Kevin had 5 and 3 throws, respectively (only possibility with 2,3,4,5, and 6 throws). By elimination, Carrie threw the ball 4 times.

In summary: (name of the child, grade, number of throws, taunt) Carrie, fourth, 4 throws, 'Yo, kid, those freckles or have you got

the measles?' Emily, third, 5 throws, 'Hey Shortie, your owner know you're off

the leash?' Kevin, second, 3 throws, 'Wow, those ears flap in the wind much,

kid?' Tommy, sixth, 2 throws, 'Whoa there kiddo, only earthlings are

allowed to play!' Uriah, fifth, 6 throws, 'Say kid, I've seen better arms than yours

on a rocking chair!'

Berry Delicious!

The food made with cranberries and blueberries wasn't pie, muffins, or cobbler, so it was the pudding and pancakes whose main ingredient was cranberries and blueberries, in some order. One day he ate pancakes, but it wasn't on Thursday (day before he gave flowers, which was the day before he visited his sister), wasn't the day he visited his grandmother, and wasn't the day he had a treat with strawberries (pancakes either with cranberries or blueberries); so on that day he presented his hostess with a box of chocolates. He didn't give his hostess a box of chocolates the day he had a treat with blueberries, so he had cranberry pancakes on the day he gave the chocolates, and on another


day he had blueberry pudding. The potted plant was a gift for his hostess exactly two days before he visited his mother, which was the day before he had a treat with raspberries; so it wasn't the day he had blueberry pudding (day after he visited his niece, which was the day after he had cobbler), the cranberry pancakes, muffins or pie, so the day he presented his hostess with the potted plant was the day he had cobbler. Thus, on four consecutive days he gave the potted plant and ate cobbler, visited his niece, visited his mother where he ate blueberry pudding, and ate a treat with raspberries. Since he didn't have blueberry pudding on Thursday, then it was on Monday he gave the potted plant and ate cobbler, on Tuesday he visited his niece, on Wednesday he visited his mother and ate blueberry pudding, and on Thursday he had raspberries.

He gave chocolates and had cranberry pancakes the day before he gave flowers, which was the day before he visited his sister; but the only possibility is that he gave chocolates on Tuesday to his niece and she served him cranberry pancakes, and on Wednesday he gave his mother flowers and she served him blueberry pudding, and on Thursday he visited his sister and she served him a treat with raspberries. He didn't have pie at his sister's, so she is one who served him muffins made with raspberries. By elimination, he had the pie on Friday, but not at his grandmother's house; so that was at his aunt's house and it was his grandmother who got the potted plant and served him cobbler on Monday. The cobbler he got on Monday at his grandmother's didn't have strawberries, so that was blackberry cobbler, and his aunt served him strawberry pie on Friday. Stefano didn't give his aunt a book, so that was his gift to his sister on Thursday, and by elimination, his aunt received the perfume (introduction).

In summary: (day of the week, relative, gift, treat) Monday, grandmother, potted plant, blackberry cobbler Tuesday, niece, chocolates, cranberry pancakes Wednesday, mother, flowers, blueberry pudding


Thursday, sister, book, raspberry muffins Friday, aunt, perfume, strawberry pie

Unique Friends

The five friends are a man nicknamed Jinx, man in DE, man in N.H., man who lives in windmill, and man who lives in old barn. Also, the five are a man nicknamed Jinx, a man nicknamed Whistler, man lives in VT, man lives in CT, and a man who lives in a tree house. Thus, man nicknamed Jinx lives in either the cave or the boxcar, and Jinx doesn't live in DE or N.H. or VT or CT so he lives in N.J. Since Jinx lives in N.J., the man who lives in the windmill and the one who lives in the old bam are the men who live in VT and CT, in some order. Thus, the man nicknamed Whistler and the man who lives in the tree house live in DE and N.H., in some order. Neither the man nicknamed Torpedo nor the one nicknamed Egghead live in the tree house, so the one who lives in the tree house must be the one nicknamed Marbles; thus, Whistler and Marbles live in DE and N.H., in some order, and Torpedo and Egghead live in VT and CT, in some order. From the above, Jinx lives in either the cave or the boxcar, but not the cave (either Aloysius or Dedrick, neither of whom is nicknamed Jinx), so Jinx from N.J. lives in a boxcar. Therefore, the men from DE and N.H. (Whistler and Marbles, in some order) live in the cave and tree house, in some order, but Whistler does not live in the tree house, so he lives in the cave, and Marbles resides in the tree house, and that leaves Torpedo and Egghead as the ones who live in the windmill and the old barn, in some order; but, Egghead does not live in the windmill, so he lives in the old barn and Torpedo lives in the windmill. Torpedo and Egghead live in VT and CT, in some order (since we've determined Jinx lives in N.J.), so neither of them is Aloysius or Dedrick or Ebenezer; they are Baldric and Cathimor, in some order. Since Jinx lives in a boxcar (from above) then he only can be Ebenezer. The man who lives in the cave is not from DE, so he is the one from N.H. and the one from


DE is the one who lives in the tree house (Marbles). Baldric is not the one who lives in VT, so he is the one who lives in CT and it is Cathimor who lives in VT; and Baldric (CT, either Torpedo or Egghead) is not nicknamed Egghead, so Baldric is nicknamed Torpedo and lives in the windmill, and it is Cathimor who is known as Egghead and lives in the old barn. Dedrick is not nicknamed Marbles, so Dedrick is the one nicknamed Whistler who lives in a cave in N.H., and it is Aloysius who is nicknamed Marbles who lives in a tree house in DE.

In summary: (first name, nickname, dwelling, state) Aloysius, Marbles, tree house, Delaware Baldric, Torpedo, windmill, Connecticut Cathimor, Egghead, old barn, Vermont Dedrick. Whistler, cave. New Hampshire Ebenezer, Jinx, boxcar. New Jersey

Strange Dreams

He dined out Monday through Friday and paid $8.00, $8.50, $9.00, $ 10.00, or $ 12.00 for meals (introduction). He didn't dine at the French restaurant the night he dreamt of the Cracker Jacks or the night he paid only $8.00 for a meal (least amount), or on Thursday night, which was the night he dreamt he was swimming in bowl of soup, so the French restaurant must be the one he gave a five-star rating. So the restaurants were the Chinese restaurant, the French restaurant, the one he dined at on Thursday night, the one he paid $8.00 for his meal, and the one where later he dreamt of Cracker Jacks. One of those five that his meal cost $12.00 (most) wasn't the one he dined at on Thursday night, the night he dreamt of the Cracker Jacks, or the French restaurant, so he paid $ 12.00 to dine at the Chinese restaurant. The restaurant he dined at the night he spend $8.00 for a meal wasn't the Hungarian restaurant or the German restaurant, so it was the Swedish restaurant where he paid only $8.00. The restaurant where he paid $8.50 for his meal wasn't the Hungarian restaurant (more than two others) or the


French restaurant, so it was the German restaurant. Since he paid $8.50 for his meal at the German restaurant, than the night he paid $8.00 was the night he dreamt that everyone's clothes were too small. He dreamt he was swimming in a bowl of soup on Thursday night, which wasn't the night he dined at the Hungarian restaurant, the Chinese or the French restaurant, so it was the German restaurant ($8.50). This leaves the night he dreamt of Cracker Jacks as the night he dined at the Hungarian restaurant (from above).

We've determined his meal at the Chinese restaurant was $12.00, the one at the German restaurant on Thursday was $8.50 (dreamt he was swimming in bowl of soup), and the one at the Swedish restaurant was $8.00 (dreamt that everyone's clothes were too small). The night he dreamt he was stuck to his chair with super glue was the night he gave the restaurant a three-star rating, so that could only be the Chinese restaurant, so by elimination, the night he dined at the French restaurant was the night he dreamt no one had any silverware, which was three stars more than the rating he gave on Wednesday night; so on Wednesday, Stefano gave a two-star rating. He gave a four-star rating on Tuesday night. Therefore, on Thursday night when he dined at the German cafe was the night he gave only a one-star rating. This leaves the two-star rating (Wednesday) and the four-star rating (Tuesday) for the nights he dined at the Swedish and Hungarian restaurants, in some order, and he paid $9.00 and $10.00 for his meals at the French restaurant and the Hungarian restaurant, in some order. He rated the German restaurant one star lower than the one where he dreamed everyone's clothes were too small so the Swedish restaurant got two stars and the Hungarian restaurant four stars. He didn't dine at the French restaurant on Monday night, so he dined at the Chinese one on Monday night and the French one on Friday night. He dined at the Swedish restaurant later in the week, than the night he paid $9.00 for a meal; so he paid $9.00 for his meal at the Hungarian restaurant, and by elimination, he paid $ 10.00 for his meal at the French restaurant.


In summary: (day of the week, restaurant, price, rating, dream) Monday, Chinese, $12.00, three-stars, stuck to chair Tuesday, Hungarian, $9.00, four-stars. Cracker Jacks Wednesday, Swedish, $8.00, two-stars, everyone's clothes too small Thursday, German, $8.50, one-star, swimming in a bowl of soup Friday, French, $10.00, five-star, no silverware

Aunt Agatha's Haunted House

On Tuesday, Stefano heard the crying and saw the ghost of the cook On Wednesday, Stefano heard a moaning sound. On Thursday, Stefano saw a ghost in the kitchen. He saw neither the parlour maid nor the housekeeper on Friday, and since he saw the ghost of the cook on Tuesday, then on Friday he saw either the upstairs maid or the butler. The tapping sound (night after he saw the ghost in the dining room) wasn't heard on Monday or Friday (kitchen was Thursday), so tin-tapping sound was heard on Thursday, and the ghost in the dining room was seen on Wednesday. The ghost of the upstairs maid was preceded by hiccupping, so it had to be sighted either on Monday or Friday. Butler (guest bedroom) wasn't sighted on Tuesday, Wednesday (dining room), or Thursday (kitchen), so he was sighted either on Monday or Friday; thus, the ghosts of the upstairs maid and the butler were seen on Monday and Friday, in some order, and the parlour maid and the housekeeper on Wednesday and Thursday, in some order. Thursday (kitchen, tapping) was not the night he saw the housekeeper, so that was the night he saw the parlour maid, and he saw the housekeeper on Wednesday night. Since the sighting of the ghost of the upstairs maid was preceded by a hiccupping sound, then the sighting of the ghost of the butler (Monday or Friday) was preceded by the whistling sound. The upstairs maid wasn't seen in the parlour, and since the butler was seen in the guest room, then it was the cook who was seen in the parlour, and the upstairs maid had to have been seen in the hallway. The ghost of the butler wasn't seen on Monday (seen near end


of week) so the ghost of the butler (guest bedroom, whistling) was seen on Friday, and the upstairs maid (hiccupping, hallway) on Monday.

In summary: (day of the week, ghost, room, sound) Monday, upstairs maid, hallway, hiccupping Tuesday, cook, parlour, crying Wednesday, housekeeper, dining room, moaning Thursday, parlour maid, kitchen, tapping Friday, butler, guest bedroom, whistling

Number Tales

Here is a series of short, surprising stories and games that we ignored when they were first mentioned. These number tales do not follow any pattern; they have been picked up the way a toddler decorates her bed with anything that glitters.

Friends Appear in Pairs

Often mathematicians discover strange behaviour of numbers long after they came to be known. The behaviour I will be talking about was probably first noticed by Pythagoras. The same person who told us that in a right-angled triangle, the square of the longest side (called hypotenuse) is equal to the sum of the squares of the other two sides. Pythagoras probably discovered that two numbers could be 'friendly' to each other. According to the Greek philosopher-mathematician, two numbers are friendly to each other if the sum of their divisors equal to each other. The first pair of friendly numbers we encounter when we start counting from 1 is 220 and 284.

How do we know that they are friendly to each other? Here is how:

220 = 1 + 2 + 4 + 71 + 142; and 284 = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110

N u m b e r T a l e s 107

The search for friendly pairs continued for a very long time. A brilliant Arab mathematician, Thabit ibn Quarra, even came up with a method to find more such pairs. Applying his rule, another Arab, Ibn al-Banna discovered the second pair, 17296 and 18416. Then, for some unknown reasons, Thabit's rule was forgotten before Frenchman Pierre de Fermat chanced upon it in 1636 and rediscovered al-Banna's pair. Two years later, French philosopher Rene Descartes found the third pair—9363584 and 9437056. Since then, many such pairs have been discovered.

Of Months, Zodiac and 12

1 ,et's take a look at the number 12. You all know that there are 12 months roughly divided into four seasons. There are also 12 signs of the zodiac divided into three sets of 4 each, and 12 hours repeated through each day and night.

The number 12 can be divided by the sum of its digits (1 + 2 = 3) and also by the product of its digits (1 x 2 = 2).

Note, 12 x 12 = 144. Now reverse the digits: 21 x 21 = 441

You can apply the same rule to number 13.

13 x 13= 169; and 31 x 3 1 = 9 6 1

Multiply a few more like these and see whether they also follow this pattern.

Reversible Magic

Is 19 + 21 equal to 91 + 12? No. Is it possible to find such two-digit numbers whose sum is equal to the sum of those very numbers with interchanged positions of the digits? Think before you say no. Mathematics is a strange world. I always think thrice before jumping to any conclusion. However, when I was asked


the same question I, like a fool, said it was impossible. My young friend quickly came up with the following example:

13 + 42 + 53 + 57 + 68 + 97 = 79 + 86 + 75 + 35 + 24 + 31

The story does not end here. If you square each of the numbers on both sides and add them, you get the same result. That is,

132 + 422 + 532 + 572 + 682 + 972 = 792 + 862 + 752 + 352 + 242

+ 312

And, here comes the real surprise,

133 + 423 + 533 + 573 + 683 + 973 = 793 + 863 + 753 + 353 + 243

+ 313

Why don't you find out more numbers that behave like the above? Happy number hunting!

Magic Squares

Let's play a game of magic squares. A magic square is an arrangement of numbers in the form of a square such that the sum of the numbers in any row, any column, or full diagonal is the same. Here is an example:

4 9 2

3 5 7

8 1 6

Let's check if our example is really a magic square. Keep adding the numbers on any row, or column or diagonal and you'll get 15 as the answer. Here it goes:

4 + 9 + 2 = 4 + 5 + 6 = 8 + 5 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 9 + 5 + 1 = 2 + 7 + 6 = 15.


Number Game

When the celebrated German mathematician Karl Friedrich (iauss was only nine years old, he was asked to add all the integers from 1 to 100. He quickly added 1 to 100, 2 to 99, and so on for fifty pairs of numbers each adding to 101. His answer was: 50 x 101 =5,050.

Here is an interesting twist to the Gauss' method. Let us find the sum of all the digits in the integers from 1 to 1,000,000,000. We will be adding all the digits in all the numbers, not all the numbers themselves. The task might look daunting, but it is actually very simple. First, group the numbers by pairs:

999,999,999 and 0; 999,999,998 and 1; 999,999,997 and 2; and so on.

There are half a billion or 500,000,000 pairs, and the sum of the digits in each pair is 81 (check it for yourself)- Now, the digits in the unpaired number, 1,000,000,000, add up to 1. So,

(500,000,000 x 81) + 1 = 40,500,000,001

Afraid of Multiplication? Follow the Russians

If you did not know how to multiply or subtract, and you were asked to find the product of 27 and 35, how would you go about it? Russian peasants found an interesting way to overcome the problem.

To multiply 27 by 35, write the numbers at the top of two columns: choose one column and halve the number again and again, ignoring any remainders, until 1 is reached. Now double the other number as many times. Here we go:


27 35

13 70

6 140 X

3 280

1 560

Now, cross out the numbers in the second column that are opposite to an even number in the first. Add the rest of the numbers in the second column to get the answer.

So we have, 35 + 70 + 280 + 560 = 945.

Prime Words

Prime numbers are the trickiest customers of the number world. The more we dig them the more we get surprised by their strange beauty. No wonder prime numbers are difficult to tame. Here are a couple of interesting facts that G.L. Honaker, Jr. discovered about palindromic Sophie Germaine Primes—191 and 383.

If we let A = 1. B = 2, C = 3, and so on till Z = 26 and add all the letters in the expression 'palindromes are fun ' , we get:

P + A + L + I + N + D + R + 0 + M + E + S + A + R + E + F + U + N = 191

Also, 15551 (the smallest palindromic prime producing another in this way) when written in words:

F + I + F + T + E + E + N + T + H + O + U + S + A + N + D + F + I + V + E + H + U + N + D + R + E + D + F + I + F + T + Y + O + N + E = 383


Two Beautiful Squares

16 = 42

1 6 - 1 = 15 and 4 - 1 = 3 Now, if we insert 15 in the centre of 16, and 3 at the left of 4,

the square relation remains unaltered. And this can be continued infinitely by inserting 15 in the

centre of the resulting number and placing 3 at the left of the resultant number on the right hand side (in each step) as shown below.

16 = 42

1156 = 342

111556 = 3342

11115556 = 33342

And it can be continued infinitely. Before proving this, let us write a number in a particular fashion:

11115556 as 14536, and, 11115556= I 4 x l 0 4 + 5 3 x 10 + 6

and, we will write 99 as 92

Since 9 9 = 1 0 0 - 1 = 1 0 2 - 1 ; 9 2 = 102 - 1

Similarly, for 99999.. . V numbers, 9n = 10" - 1

Similarly, a number of this type consisting of '2n' digits can be

written as ln5n_|6, T h e r e f o r e , l n 5 n . 1 6 , = l n x l 0 " + 5 n . 1 x l 0 + 6

= l n x 10n + 5 n + 1 = (9 x 10" + 5n x 9 + 9)/9 = {("l0" - 1) x 10" + 5 x (10" - 1) + 9}/9 = (102" + 4 x 10" + 4)/9 = (10" + 2)2/9 = (9n + 1 + 2)2/9


= {(9n + 3)/3}2

= ( 3 n + l ) 2

= 333 42

Another number, 49 also shows these characteristics. (Biswajit Ganguly)

The House of Santa Claus

The house of Santa Claus is a very old German drawing game for small children. You have to draw a house in one line, without lifting your pencil while drawing. You must also not repeat a line. As you draw, you have to say: Das istdas Haus des Nikolaus, or keeping the rhyme in mind 'This is the house of Santa Claus'. You must speak a syllable of this sentence to every straight line. In how many ways can you draw such a house? Here are some examples:

The computer (C64 nostalgia) checked all numbers between 111,111,111 and 155,555,552, to find a solution. It found 44 possibilities of drawing a house in this manner. Here are the solutions:


1 , 2 , 3 , 1 , 4 , 3 , 5 , 4 , 2 1 , 3 , 4 , 5 , 3 , 2 , 4 , 1 , 2 1 , 2 , 3 , 5 , 4 , 3 , 1 , 4 , 2 1 , 3 , 5 , 4 , 3 , 2 , 1 , 4 , 2 1 , 2 , 4 , 5 , 3 , 1 , 4 , 3 , 2 1 , 4 , 2 , 3 , 5 , 4 , 3 , 1 , 2 1 , 3 , 2 , 4 , 5 , 3 , 4 , 1 , 2 1 , 4 , 3 , 5 , 4 , 2 , 1 , 3 , 2 1 , 3 , 4 , 2 , 3 , 5 , 4 , 1 , 2 1 , 4 , 5 , 3 , 2 , 4 , 3 , 1 , 2 1 , 3 , 5 , 4 , 2 , 1 , 4 , 3 , 2 1 , 2 , 3 , 4 , 5 , 3 , 1 , 4 , 2 1 , 4 , 2 , 1 , 3 , 5 , 4 , 3 , 2 1 , 2 , 4 , 3 , 1 , 4 , 5 , 3 , 2 1 , 4 , 3 , 2 , 1 , 3 , 5 , 4 , 2 1 , 3 , 2 , 1 , 4 , 5 , 3 , 4 , 2 1 , 4 , 5 , 3 , 1 , 2 , 4 , 3 , 2 1 , 3 , 4 , 2 , 1 , 4 , 5 , 3 , 2 1 , 2 , 3 , 1 , 4 , 5 , 3 , 4 , 2 1 , 3 , 5 , 4 , 1 , 2 , 3 , 4 , 2 1 , 2 . 4 , 1 , 3 , 4 , 5 , 3 , 2 1.'3.:5,4, 3 , 2 , 4 , 1 , 2 1 2 , 4 , 5 , 3 , 4 , 1 , 3 , 2 1 ,4 ,3 , 1 , 2 , 3 , 5 , 4 , 2 1 , 3 , 4 , 1 , 2 , 3 , 5 , 4 , 2 1 , 4 , 3 , 5 , 4 , 2 , 3 , 1 , 2 1 , 3 , 4 , 5 , 3 , 2 , 1 , 4 , 2 1 , 4 , 5 , 3 , 4 , 2 , , 3 , 2 1 , 3 , 5 , 4 , 2 , 3 , 4 , 1 , 2 1 , 2 , 3 , 5 , 4 , 1 , 3 , 4 , 2 1 . 4 , 2 , 3 , 4 , 5 , 3 , 1 , 2 1 , 2 , 4 , 3 , 5 , 4 , 1 , 3 , 2 1 , 4 , 3 . 2 , 4 , 5 , 3 , 1 , 2 1 , 3 , 2 , 4 , 3 , 5 , 4 , 1 , 2 1. 4, 5, S, 2, 1 , 3 , 4 , 2 1 , 3 , 5 , 4 , 1 , 2 , 4 , 3 , 2 1 , 2 . 3 , 4 , 1 , 3 , 5 , 4 . 2 1 , 4 , 2 , 1 , 3 , 4 , 5 , 3 , 2 1 ,2 ,4 , 1 , 3 , 5 , 4 , 3 , 2 1 ,4 ,3 , 1 , 2 , 4 , 5 , 3 , 2 1 , 3 , 2 , 1 , 4 , 3 , 5 , 4 , 2 1 , 4 , 5 , 3 , 1 , 2 , 3 , 4 , 2 1 , 3 , 4 , 1 , 2 , 4 , 5 , 3 , 2 1 , 4 , 5 , 3 , 4 , 2 , 3 , 1 , 2

Note that all sequences start in 1 and end in 2. You can conclude that there are also 44 houses starting in 2 and ending in 1. They are the same because of their symmetry. (Jtirgen Koller)

The Continuing Saga of Big Burly Bootless and Sleepy Sloppy: The Dust Solution

Prelude The tunnel ride was long. Darkness inside the tunnel clouded his thoughts. Big Burly Bootless wanted to open the windows.


His partner, Sleepy Sloppy, protests. 'Too windy here,' he says. 'I 've lost my woollen scarf,' Sloppy mumbles feebly. Bootless spews venom under his breath, ' . . .as if he needs woollens, the hairy body can handle even the North Pole winter; woollen... huh!'

They have been driving for days now, stopping only for hydrogen and playing Butch Cassidy & Sundance Kid on the windscreen over and over again. Despite the mutual hatred, Sloppy and Bootless have two common passions—stealing ancient motor parts and watching the Paul Newman-Robert Redford classic.

It was winter 2167, when Sloppy met Bootless for the first time at Little Jollie's gene transfer centre. While searching mobile male databases, Dr Dimly Dust came across Bootless' profile. He was offering his genes for 'slow-moving nitwits'. Three-year-old Sloppy was a perfect match.

Dr Dust was pleasantly surprised. 'The best match in less than three nanoseconds! That beats the last record at this centre by 27 nanoseconds, quite remarkable,' the expert had commented. It took two whole minutes to transfer Bootless' fast-thinking genes into Sloppy's slow-moving grey cells. It was a long medical procedure. Bootless had insisted on the long-discarded bacterial transfer of his genes. By the end of the procedure, even Dr Dust's nano-assistant Twink was tired.

Dr Dust's Solution

Despite their apparent bonhomie, Sloppy and Bootless continuously fought over which motor part to steal. Dr Dust's message-reading glasses often got blurred with complaint messages from the warring duo. One day, after consulting Twink, Dr Dust found the solution. He invented a game and called it Colonize. It is a game played on a 2(XX) x 2000 square grid with


numbers. Twink was right, the number of messages from Sloppy and Bootless reduced exponentially.

Here is a version of the game that engaged the gene partners

for hours:

Colonize (you can play the game without laser guides

and sound zappers)

Table 17

0 11 80 27 32 44 3 69 0

30 5 15 72 21 81 54 63 47

61 75 22 40 55 7 36 51 18

9 68 43 52 78 26 10 17 33

70 38 3 16 49 62 73 23 6

45 8 79 31 67 14 29 60 56

19 34 57 77 2 35 66 41 28

24 46 71 65 53 20 12 39 1

0 25 37 4 13 50 48 6 0

At this instance, we'll use the above grid to play. However, you can draw >our own grid and put any number in any box.

Rules: . All players start from the four corners of a square, the

youngest to move first. . Announce the number of steps you will use to reach a


particular number, but do not announce the number you want to 'colonize'. Write down the steps on a writing pad and keep it upside down. Let your opponent announce his or her steps. After all the announcements are made, you can move (if you are the youngest). You will have to show your movements to your opponents when challenged. No diagonal movements are allowed in this game. A step means moving from one square to an adjacent square. Once you colonize a square, your opponent cannot move in there. However, you can allow your opponent to use one or more of your squares in exchange of an equal number of squares during your move. From the start of the game to five moves, a player has to use a minimum of five steps to colonize a square. For the next seven moves, one has to use a minimum of four steps. And for the rest of the game, it is at least three steps. If all four players want to colonize the same square, the player starting from the least number square gets it. All other players go back to their respective starting points. The game ends when there are no more squares left to colonize. Prime Steps: Announce when you are going to colonize a prime number. If you can successfully colonize it, you can escape a penalty. Penalty: If you cannot land on the prime-numbered square of your choice, then for single-digit primes, move the same number of squares to the left along the row in which your choice of square is placed (starting from the prime square). If the number exceeds the number of squares in that row, go down one row and keep moving to the right. For the first and the eighth row, go back to your starting point. In case


Of multiple-digit primes, keep adding the digits till you reach a single-digit number and start moving to your left. (Note: 1 and 2 are excluded from the list of primes.)

. W h o w i n s ? T h e player with the maximum number of 'colonize-

d' squares!

Beyond the Game One might try and answer an interesting question. For an n x n grid with the same rules being applied, will the first mover always win the game?

Puzzlemania I

Carrier Pigeons

The only place in India where the police department used carrier pigeons to deliver messages is Orissa. A carrier pigeon departs from Bhubaneswar for Cuttack at the same time as another carrier pigeon departs from Cuttack for Bhubaneswar. Both pigeons fly at constant speeds, although different from each other. They cross paths 2x kilometres from Bhubaneswar. After each arrives at their destination, they immediately turn around, going back and forth without breaks. They cross paths the second time x kilometres from Cuttack. Where will they cross paths the third time?

Free Script for a Bollywood Blockbuster

Here is a fantastic recipe for a Bollywood blockbuster. Imagine you are directing the film and call it Dil Mangta Hai. Here is the plot: Four men, Shah Rukh, Aamir, Vivek and Manoj, and four women, Rekha, Tabu, Preity and Nandita, are travellers caught in a no-man's land for several years. With very little to do, eventually each person falls in love with one person and is loved by one person. The audience knows this:

P u z z l e m a n i a II 119

• Shah Rukh loves the girl who is in love with Aamir. | • Vivek loves the girl who loves the man who loves Tabu.

• Manoj loves the girl who loves the man who loves Preity. | • Rekha does not love Manoj. I • Nandita loves a man who does not love Rekha.

Alter several songs—use A.R. Rahman, please—the audience is told who loves whom. If you want a producer for this film, first untangle the four-by-four romantic knots.

Ragging Indian Style

l ite Scene: The introduction of new students to ways of a college hostel. On the first day, new boarders to perform an odd opening day ceremony:

There are 500 lockers and 500 new boarders in the hostel. The hostel bully asks the first boarder to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the 500th student, how many lockers are open?

A Very Old Puzzle

It was Sleepy Sloppy's first day at school. The teacher suggested that it would be a good idea for each child to meet every other child in the class. The teacher said, 'When you meet, please shake hands and introduce yourself by name.' If there were 20 children in the class, how many total handshakes were there?


Hot Pursuit

FB I Agent Alice is hot on the trail of computer hacker Bob, who is htding in one of seventeen caves. The caves form a linear array, and every night, Bob moves from the cave he is in to one of the caves on either side of it. Alice can search two caves each day, with no restrictions on her choice.

For example, if Alice searches caves 1 and 2, then 2 and 3 and so on till caves 16 and 17, then she is certain to catch Bob though it might take her sixteen days.

But can you find out the shortest time in which Alice can be guaranteed of catching Bob? (Guilford-Wagon)

Egg Dropping

Suppose we wish to know which windows in a thirty-six-storey building are safe to drop eggs from, and which will cause the eggs to break on landing. We make a few assumptions: • An egg that survives a fall can be used again. • A broken egg must be discarded. • The effect of a fall is the same for all eggs. • If an egg breaks when dropped, then it would break if dropped

from a higher window.

• If an egg survives a fall then it would survive a shorter fall • It is not ruled out that the first-floor windows break eggs

nor is it ruled out that the thirty-sixth-floor windows do not' cause an egg to break.

If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried out in only one way. Drop the egg from the first-floor window; if it survives drop ,t from the second floor window. Continue upwards until


it breaks. In the worst case, this method may require thirty-six droppings. Now suppose two eggs are available. What is the least number of egg-droppings that is guaranteed to work in all cases?

Autorickshaw Drive

To move goods, our honourable roadways ministers have built more than 50,000 km of roads. Now, imagine driving those miles in the first petrol-driven vehicle that has only three wheels—an autorickshaw—and could reach a top speed of about 10 km an hour. For safety's sake, all autorickshaws carry a spare tyre. As you drive the 50,000 km, you rotate the spare with the other tyres so that all four tyres get the same amount of wear. Can you figure out how many miles of wear each tyre accumulates?

By Gauss, What's the Order?

There are eight volumes in the complete works of Carl F. Gauss. They are initially arranged in an arbitrary order. A student wants to place the volumes in the correct order, i.e. 1, 2, 3 ,4 , 5, 6, 7 and 8. However, she is only allowed the following two operations. She can take a volume from either the third position (counting from the left) or the eighth position and place it first. Can she achieve her goal? (C. Gao)

Free Tripper

At a movie theatre, the manager announces that they will give a free ticket to the first person in line whose birthday is the same


as someone who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, that birthdays are distributed randomly throughout the year, etc., what position in line gives you the greatest chance of being the first duplicate birthday person? (David Karr)

Treasure Hunt

A school hallway has a long row of lockers. Every sixth locker contains a package of chewing gum, every eighth locker contains a hockey stick, and every ninth locker contains a mirror. Which is the first locker to contain all three items?

Weakest Runner-Up

This is one of those puzzles born out of love for 'The Beautiful Game'—football. In a World Cup, there are thirty-two teams numbered Kl , K2, K3.. . K32. They are randomly distributed in eight groups—A, B, C, D, E, F, G and H. The technical committee of FIFA has already assessed the teams, and according to them, Kl is the weakest and K32 is the strongest. The teams in between are as strong as their numbers on the list. The teams follow the playing format displayed in Figure 11. Where al denotes the winner of group A; b2 is the runners-up of group B; . . .gl is the winner of group G, h2 is the runners-up of group H. Now, tell us:

• The weakest team to qualify for semi-finals; and • The weakest to be the runners-up of the tournament

(Partha Chowdhury, Kolkata)


al a2 dl c2

el f2 hi g2

bl a2 cl d2

f l e2 g>

h2

Figure 11

Popcorn, Cola—It's Movie Time, Folks

A boy passing by a cinema on a bus could notice only the hours and not the minutes when four of the eight shows began. This is what he saw:

First Show: 12 hrs ...min Second Show: 13 hrs ...min

Seventh Show: 23 hrs ...min Eighth Show: 24 hrs ...min If the duration of the shows is the same and there is no time interval between any two shows, find when did the first show start ? (Shamik Chaudhury, Serampur)

Your Number Please

To settle any acrimonious debate, Prof. Serene always calls her dear friends—the numbers. Last time her friends helped settle a debate over who is a better cricketer, Sachin Tendulkar or Brian Lara.

Here is how she separated the warring factions: 'Let N be a


500-digit integer divisible by 99,' she said. 'And let N* be a number obtained by reversing the order of the digits of N.' Then, she claimed: 'N* is also divisible by 99.'And asked: 'Am I right?'

Not a Game of Chess

I was sitting before my chessboard, pondering a combination of moves. At my side were my daughter and son. The girl was busy with her homework, which consisted of some exercises in long division, but she was handicapped by her naughty brother, who kept covering her figures with chessmen. As I looked up, only two digits remained visible (Figure 12). Can you calculate the missing digits?

T a 4 8 & n fiCrn A & i i & & &

X X I 1

i l l

1 1 1

1 Figure 12

A Walk in the Garden

The plan of my future garden (Figure 13) has the four junctions A, B, C and D, and six roads AB, AC, AD, BC, BD and CD. In the plan, four roads—AB, BC, AC and CD—have grass verges and two roads—AD and BD—have flowers verges. Every morning


I imagine myself standing at the mid-point of one of the roads and facing one direction of the path. I ask my daughter, who is not aware where I'm standing or which way I 'm facing, to give me instructions for a walk around my park: she gives me three words in order and each is either 'left ' or 'right'. For example, she tells me either left-left-left or right-left-right and so on. I lollow her instructions in turn at the three junctions I come to «>n my walk. When I follow her last instruction, I stop halfway along the road I am on and tell her whether I have a flower verge or a grass verge. Then she gives me another set of instructions for my walk and I repeat the exercise. We continue this way till she can deduce what my starting point was and which direction I faced. She always gets the answer by giving me the minimum number of walks. What is the minimum number of walks?

A

Figure 13

Marble Story

Bulbul and Chhotu love to play marbles. They often skip school to be able to do so. One afternoon, Father Frank caught them


and took all their marbles. After he finished counting, he told the boys, 'The number of marbles you have is very interesting. It is a unique number.' Since the boys were not aware of the number of marbles, Father Frank told them how to calculate it. 'It is a four-digit number. And the first digit raised to the power of the second digit multiplied by the third digit raised to the power of the fourth digit is equal to the number. Also, the same digit appears twice in the number and none of the digits is zero.' How many marbles did Father Frank take from Bulbul and Chhotu?

Wordwallah's 'Words Mall'

Woody Wordwallah has nothing to offer but words. The old man from the city of Mumbai has recently unveiled his latest game 'Words Mall' at his Word Party to celebrate his ninety-ninth birthday. He invited ninety-nine friends and told them, 'I have created a new game. Take a close look at the board.'

W 0 R D S M A L L

Then Woody covered the board with a black cloth, showing only the squares drawn in white lines and not the letters. 'I have exactly one of each letter in the form of clay tablets of the word "words mall" in my bag,' he told his friends. 'One by one, each of you will get to play the game. Just put your hand inside the bag, take out the tablets, one at a time and in random order, and without looking, place them upside down on the black cloth, in order starting from the left,' Woody announced the rules. 'However,' he paused, 'to play this game you have to pay an entry fee of 1 silver coin. And for each letter tablet that matches with the letters of


"words mall", I will pay you 1 silver coin.' If all of Woody's friends play the game, how much does the old man expect to win or lose?

Here Comes the Hangman

X, Y, Z are three royal prisoners, of which two are to be hanged. Finally, on the day of reckoning, the guard tells them that the king has made his decision on which one of them is to be released. X asks the guard who is the lucky one. Instead of giving him a direct answer, the guard tells him that Y is to be hanged. Do the odds of X being released change due to the guard's statement? (Soumak Chatterjee)

Ball Game, Sen Style

On a charity mission, Sushmita Sen went to South Africa and was asked by the students to participate in a game. Sushmita and all the students stood in a line and were asked to draw balls from a container. There were 32 boys with Sushmita and 33 balls, of which one was red. Whoever would draw the red ball would win the prize. Every boy was presented a ball by the machine randomly. Since Sushmita was a guest, she was given the opportunity to choose her own position in the line. Where should Sushmita stand in the line to maximize her chances of getting the prize? The game stops as and when the red ball is drawn! (Soumak Chatterjee)

One for the Dog

Five friends and their pet dog were spending a night in a hotel. They had a large packet of biscuits and had agreed to open it the


next morning. They were too tired and soon were sleeping. One of them woke up at midnight with hunger and opened the packet of biscuits. He divided the biscuits into five equal lots and found one biscuit extra. He ate his share and gave the extra one to the dog. He neatly sealed the packet with four equal lots of biscuits and went to sleep. After a few minutes, another friend woke up. He did the same as the first one—divided the biscuits into five equal lots—and found one biscuit extra. He ate his share, gave the extra biscuit to the dog, sealed the packet and went to sleep. The other three woke up at different times and did what their first friend had done. All of them found one extra biscuit while dividing the biscuits into five equal lots. The question is: what is the least number of biscuits the five had with them before they went to sleep?

Calculate, Do Not Guess

A high-school maths teacher chose four of his best students to participate in an extra-credit project. He told them he had chosen a long division calculation, expressed as N = D x Q + R, in which N, the dividend, was a four-digit perfect square; D, the divisor, was a three-digit perfect square; Q, the quotient, is defined as the integral part of N/D, and R, the remainder, is defined as N - (D x Q). Each of the four students was given the value of one of the four quantities, N, D, Q, R, and was told to calculate the three others. Each student knew which quantity each of the others was given, but not its value. The teacher told them to begin working but not to speak until he questioned them. At different intervals of time, the students submitted their answers, all of which were correct. Later, the teacher told a friend about the test and asked if he could determine the four numbers. After working on it for a day, the friend complained, 'I found several


possible solutions. I need more information. Could you tell me which student knew which quantity?' The teacher replied, That's far too much information! In fact if I just told you which

quantity the last student knew you could find the numbers/ What were the four numbers?

Burrowing for Gold

My dreamland looks like a solid tetrahedron; each of its four triangles is an equilateral one, with sides of 30,000 m each, covered with lush green grass. Every night, I start at the mid-point of one edge and burrow in a straight line through the dreamland, and emerge at the mid-point of the opposite edge to get a pot of gold coins. While I am busy, my doppelganger starts to walk from the same point where I did and takes the shortest route on the surface of the dreamland to the point where I emerge. Competition! No problem, he is only my doppelganger. Tell me, how far does my doppelganger walk?

A Quick Game with Your Friends

You asked each of the four children to think of a four-digit number. 'Now, transfer the last digit to the front and add the new number to the old one. For example, 5432 + 2543 = 7975. Now, tell me the results.'

Bappa: 2348 Motu: 7847 Jaypee: 11847 Dyuti: 9846

'Everyone except one is wrong,' you told the bunch. Who was it and how did you know?


Who Scored More?

In a class test, my elder son Arkadev got marks equal to the remainder when 7123456789 is divided by 11. In another test, my younger son Rudradev got marks equal to the remainder when 7123456789 j s d i v i d e d 5 y 1 3 W h o g Q t m Q r e m a r k s ? (prabir Basak, Burnpur)

PM Wins Thirteenth Round

Pollster Manipulator has been winning elections ever since his father's death. Hoodland's most famous statesman believes in ultimate democracy. Detractors say Manipulator's popularity is nothing but hoodwinking the people. However, elections are held every six months and no one has even come close, let alone beating, the election champ. In the last elections, his thirteenth, Manipulator announced that people would be using small pith balls in the box that had the candidates' name inscribed on it. However, there was a twist; the balls should form a pyramidal stack. Each layer in the stack should have a square array of pith balls and each layer must have one ball less along the edge than the layer immediately beneath it (except the lowest layer). In this way, the top layer will have a single pith ball. Each pith ball is equal to one vote. Needless to say, Manipulator won the elections hands down. Only his box had the pyramidal structure as mentioned above. A specially designed box with slots came handy but that is another story. Manipulator's pyramid had 19 layers. Tell me, how many votes did he get?

Santa's Order

Santa always leaves plans for his elves to determine the order in which the reindeers will pull his sleigh. This year, for the


I uropean leg of his journey, his elves are working to the following schedule, which will form a single line of nine reindeers:

C !omet behind Rudolph, Prancer and Cupid; Blitzen behind Cupid and in front of Donder, Vixen and Dancer; Cupid in front of Comet, Blitzen and Vixen; Donder behind Vixen, Dasher and Prancer; Rudolph behind Prancer and in front of Donder, Dancer and Dasher; Vixen in front of Dancer and Comet; Dancer behind I )onder, Rudolph and Blitzen; Prancer in front of Cupid, Donder and Blitzen; Dasher behind Prancer and in front of Vixen, Dancer and Blitzen; Donder behind Comet and Cupid; Cupid in front of Rudolph and Dancer; and Vixen behind Rudolph, Prancer and Dasher.

Can you help the elves work out the order of the reindeers?

Trick Unmasked

A company has the policy that any employee's birthday is a holiday for the entire company. How many people should the company employ if the expected value of the total number of 'people-days' (the product of the number of employees and the number of days worked) is to be maximized? Answer the question assuming that there are 365 days in a year, that each day is equally likely to be a birthday, and that the employees have no days off except for the birthday /holidays (i.e. no weekends off). (Chance magazine)

Thrice as Big

On a safari to Kenya, the Indian cricket team saw many lions and many more other animals. The inquisitive members of the team asked their host what could be the number of lions in the country. The host was not aware of the exact number so he


decided to sound a bit vague. 'Surprisingly, the number of lions is the same as the smallest positive integer whose first digit is 1 and which has the property that if this digit is transferred to the end of the number, the resulting number is 3 times as large as the original,' he told the cricketers. What is the number?

Take Your Pencil for a Walk

Here is something that Lewis Carroll gave us. Draw the figure shown below, without lifting your pencil from the paper, without retracing any edge and without crossing your own path. How many such paths are possible?

Figure 14

The Challenge

Getting bored on a train journey to Chennai, Bibi and Bubu ask Uncle Numboo to play with them. Uncle Numboo was busy reading, so he decided to give the children a difficult puzzle. 'I have two numbers, M and N. Interestingly, M2 - N2 = A3. If I say A is a positive integer, prove to me that the equation always has proper fraction solutions for M and N.'

Bubu and Bibi were busy checking the results. After a while,


they ran to Uncle Numboo and said, 'Don't lie.' Why did the children say so?

The Second Visit

It's restful sitting in Tom's cosy den, talking quietly and sipping a glass of his Madeira. I was there one Sunday and we had the usual business of his clock. When the radio gave the time at the hour, the Ormolu antique was exactly three minutes slow. 'It loses seven minutes every hour,' my old friend told me, as he had done so many times before. 'No more and no less, but I've gotten used to it that way.' When I spent a second evening with him later that same month, I remarked on the fact that the clock was dead right by radio time at the hour. It was rather late in the evening, but Tom assured me that his treasure had not been adjusted nor fixed since my last visit. What day of the week was the second visit?

Know Your Dish

Five people go to a Lebanese restaurant. Not being familiar with such food, they do not recognize any of the names for the dishes. Each orders one dish, not necessarily distinct. The waiter brings the dishes and places them in the middle, without saying which is which. At this point, they may be able to deduce some. For example, if two people ordered the same item, and everyone else ordered different dishes, then the item of which two plates arrive must be the one of which two were ordered. They return to the restaurant two more times, following the same drill, though with different orders. After three meals, they have eaten all nine items on the menu, and can tell which is which. What pattern of ordering fulfils this?


Game Show Winners

Five families were competing on a television game show, and the objective was to match five famous faces with five famous names. Each family made the following guesses:

• The Addams family listed Otto, Roebling, Steinmetz, Tesla, and Westinghouse

• The Brady family listed Tesla, Westinghouse, Otto, Steinmetz, and Roebling

• The Cunningham family listed Roebling, Tesla, Steinmetz, Westinghouse, and Otto

• The Flintstone family listed Tesla, Roebling, Otto, Steinmetz, and Westinghouse

• The Jetson family listed Tesla, Westinghouse, Steinmetz, Otto, and Roebling

No two families got the same number of correct matches. Which family went home with the first prize?

Get Rich Quick

Stephen had a dream that he was in a cave during medieval times. Before him is a single row of hundred gemstones. WTiile checking for secret traps, he discovers a trip wire, and suddenly hundred evil gnomes appear. The evil gnomes cast Curses of Negation. If a curse of negation is cast upon a not-cursed item, that item becomes cursed and will cause instant death if touched by a human. But if it is cast upon an already cursed item, that item returns to its not-cursed state. The first evil gnome casts a curse of negation on every gem. The second evil gnome casts a curse of negation on every second gem. The third evil gnome casts a curse of negation on every third gem. This pattern continues until the last evil gnome casts a curse of negation on only the


hundredth gem. At that point, all the gnomes return to their secret chamber, and Stephen is free to take any gems that he chooses. If all the gems were not cursed prior to the arrival of the evil gnomes, how many gems can he take away safely?

Getting Better on the ELO Ratings

After the end of a one-round chess tournament with five players A, B, C, D and E, who played each other once and finished in this listed order, the players share their impressions.

• 'I am the only one who finished without a single loss,' said B. • 'I am the only one who did not win a single game!' said E.

Given that a winner got a full point, for a draw each opponent got half a point and that the ranking could be decided by only looking at the points, can you reconstruct the tournament table?

Fundoo Island

()n his last visit to his family island, Funtrap Fundoo found new roads were being built. There are six small towns on the Fundoo Island. However, the engineers there faced a problem: the six town mayors had decided that 'direct' roads should interconnect the towns. In other words, roads on the island do neither cross nor branch outside the towns. With the number of vehicles growing every day, the engineers must build the maximum number of such roads without disrespecting mayors' decision. How many such roads (at most) can they build on the island?

Jousting Knights

The following incident took place at a jousting tournament. The Red and the Black Knights stand at opposite ends of the


arena. They simultaneously spur their horses and charge. The horse of the Red Knight is slower and so they meet 60m from the nearer end of the arena, missing each other. They continue galloping at full speed and turn their horses at the end of the arena (which takes equal time for both of them). Charging again, this time they meet 50m from the nearer end and the Red Knight manages to lift his opponent from the saddle.

If both knights rode at constant speed, how long is the arena?

Postage Stamp Geometry

Fold a block of eight postage stamps, as indicated below, along the perforations and without tearing, so as to form a single pile with the stamps numbered consecutively from top to bottom.

i 1 I 1 1 | Stamp 6 1 Stamp 3 1 Stamp 5 ' Stamp 4 |

| 1 1 i | | Stamp 7 I Stamp 2 ' Stamp 8 ' Stamp 1 |

I L _L J I -I

Figure 15

Solutions

Carrier Pigeons

Assume that the distance between Bhubaneswar and Cuttack is 1. Call "Roomer' the pigeon that starts from Cuttack and 'Victory' the pigeon that starts from Bhubaneswar. For any given amount of time, the ratio of the distance travelled by Roomer to Victory will be the same. This ratio at the first meeting is (1 - 2x)/2x. At the second meeting


this ratio is (2 - x)/(l + x). Equating these two ratios: (1 - 2x)/2x = (2 - x)/(l + x), solving for 'x' we get, x = 1/5.

So, the first time they meet 2/5 of the way from Bhubaneswar and the second time 1/5 of the way to Cuttack. The easiest approach at this point is to just follow the two paths as the birds continue to fly and see where they meet. When Victory has travelled 2/5 of the distance between Bhubaneswar and Cuttack to a point 3x the distance from Cuttack, Roomer will have travelled 3/5 of this distance to 2x the distance from Cuttack. When Victory has travelled another 2/5. she will be exactly in Bhubaneswar. This is also where Roomer will be.

I ree Script for a Bollywood Blockbuster

l abu loves Shah Rukh who loves Preity who loves Aamir who loves Rekha who loves Vivek who loves Nandita who loves Manoj who loves Tabu.

Ragging Indian Style

I he only lockers that remain open are perfect squares (1,4,9, 16, etc.) Iiecause they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed. So the number of open lockers is the number of perfect squares less than or equal to 500. These numbers are l2, 22, 32, 42, and so on, up to 222. (232 is 529, and, therefore, out of range.)

So the answer is 22.

A Very Old Puzzle

The class has 20 children. The first child shakes hands with the other 19 children. The second child has already shaken hands with the first


child, and so has to shake hands with only the other 18 children. In this manner, the second-last child has to shake hands with only one child, and the last child has already met all the children. Thus, the number of handshakes is:

19+ 18+ 17 + ... + 2 + 1 = 190

Hot Pursuit Alice can catch Bob in 10 days.

Day Caves Searched 1 2,4 2 5,7 3 8, 10 4 11, 13 5 14, 16 6 2,4 7 5,7 8 8, 10 9 11, 13 10 14, 16

Egg Dropping

The solution uses the fact that 36 is the eighth triangular number, i.e. 36 = 1 + 2 + ... + 8. Here is an algorithm provided by I.P. Sealy (Mount Allison College):

• Drop the first egg from level 8. If it breaks, drop the second egg from 1, 2 ... 7 as possible.

• If the first egg doesn't break, drop it from level 15. If it breaks, drop the second egg from 9, 10 ... 14 as possible.


• If the first egg doesn't break, drop it from level 21. If it breaks, drop the second egg from 16, 17, 18, 19, 20 as possible.

• If the first egg doesn't break, drop it from level 26. If it breaks, drop the second egg from 22, 23, 24, 25 as possible.

• If the first egg doesn't break, drop it from level 30. If it breaks, drop the second egg from 26, 28, 29 as possible.

• If the first egg doesn't break, drop it from level 33. If it breaks, drop the second egg from 31, 32 as possible.

• If the first egg doesn't break, drop it from level 35. If it breaks, drop the second egg from 34.

• If the first egg doesn't break, drop it from level 36. If it breaks, the answer is 36 and you have a good egg left.

Note, if there were a seven-dropping solution, then the first drop must be from the seventh floor, the second drop from no higher than the thirteenth floor and so on. This would not reach the top floor.

Autorickshaw Drive

Since the four wheels of the three-wheeled car share the journey equally, simply take three-fourth of the total distance (50,000 km) and you'll get 37,500 km for each tyre.

By Gauss, What's the Order?

The answer to this question is 'yes'. First, let us observe that she can transpose any two volumes of the set without disturbing other volumes. Without loss of generality, let us assume that the order is 21345678 and she wants to transpose the first two volumes. Then she may obtain the following sequence:

21345678 ^ 82134567 • 18234567 • 71823456 • 87123456 • 68712345 • 56871234 • 85671234 • 48567123 • 34856712 • 83456712 • 28345671 • 12834567 • 81234567 • 78123456 • 67812345 • 56781234 • 45678123 • 34567812 • 23456781 • 12345678.


Next, observe that if she wants to transpose, say, volumes in the sixth and seventh positions, she can shift them to the first and second positions (by placing repeatedly the volume in the eighth position to the first position), perform step 1 and then shift them back to the original sixth and seventh positions. And finally, observe that by repetitive transposing pairs of volumes that are not in the correct order (i.e. 17 is the correct order whereas 71 is not) she will place all volumes in the correct order after a finite number of steps.

Free Tripper

Suppose you are the Kth person in line. Then, you win if and only if the K - 1 people ahead all have distinct birthdays and your birthday matches one of theirs.

Let A be the event that your birthday matches one of the K - 1 people ahead; and B be the event that those K -1 people all have different birthdays. Then Probability (A) = Probability (B) multiplied by Probability (AIB), where Probability (AIB) is the conditional probability of A given that B occurred.

Now let P (K) be the probability that the Kth person in line wins, Q (K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins).

Then,

P(l) + P(2) + ... + P(K - 1) + P(K) = 1 - Q (K) P (1) + P (2) + ... + P (K - 1)

= 1 - Q (K - 1) P (K) = Q (K - 1) - Q (K)

For convenience in calculation, let's set K = I + 1, and let D be the number of days in a year. Then,

Q (I - 1) - Q (I) = Q (I)*(l - 1)/(D -1 + 1) Q (I) - Q (I + 1) = Q (I)*I/D P (K) - P (K - 1) = P ( I + l ) -P ( I ) = {Q (I) - Q (I + 1)} - {Q (K - 2) - Q (K - 1)} = Q (I)* {I/D - (I - 1 )/(D -1 + 1)}


To find out where this is last positive (and next goes negative), solve x/D - (x - 1)/(D - x + 1) = 0. Multiply both sides with D*(D+ 1 - x):

(D+ 1 - x)*x - D*(x - 1) = 0, or Dx + x - x2 - Dx + D = 0, or

x2 - x - D = 0, or x = [ l ± {1 -4*(-D)}]/2

We take the positive square root = 0.5 + (D + 0.25).

Setting D = 365, desired I = x = 0.5 + (365.25) = 19.612 (approximately).

The last integer I for which the new probability is greater than the old is therefore I = 19, and so K = I + 1 = 20. You should try to be the twentieth person in line.

Treasure Hunt

Multiples of 6: 6, 12, 18, 24, 30, 36, 42. 48. 54, 60, 66, 72... Multiples of 8: 8, 16, 24, 32. 40, 48, 56, 64, 72... Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72... LCM = 72

So, locker 72 is the answer.

Weakest Runner-Up

The answer is K14 and K28. Here is a solution from Asit Kumar Sadhu, DGMS, Dhanbad.

Qualifying to semi-finals from one side involves four groups. The strongest team among the winners of two groups and the runners-up of two groups qualify for the semi-final. In other words, the strongest among the 14 teams, leaving the winners from two groups from which runners-up are playing, qualify for the semi-final. So, if we form the four groups A, B, C and D with the sixteen weakest teams in the following manner...


Group Teams A Kl, K2, K3, K4 B K9, K10, Kl 1, K15 C K12, K13, K14, K16 D K5, K6, K7, K8

.. .then, from the top left-hand corner of the given fixture, K14 qualify for semi-final as follows:

K4

Kl 1

K8

K14

Kl 1

K I 4 K I 4 K I 4 K14 (semi-finalist)

To qualify for the final match, all the eight groups are involved. Winners and runners-up of four groups will play to qualify for the final. Applying the same logic as above, K28 can qualify for the final if we form the groups as follows:

Group Teams A Kl, K2, K3, K4 B K17, K18, K19, K29

C K20, K21, K22, K30 D K5, K6, K7, K8 E K9, K10, K11.K12 F K23, K24, K25, K31

G K26, K27, K28, K32

H K13, K14, K15, K16


So, K28 can be the weakest team to be the tournament's runners-up. It will be beaten by K32 in the final.

Popcorn, Cola—It's Movie Time, Folks

Here is an elegant solution that S. Majumdar, Patna, had sent to me a lew years ago.

Let 'k' be the duration of the show in minutes and let the first show start at 12.xx hours. We assume 0 xx 60. The time difference between show'm' and show 'n' (m > n) is equal to (m - n) k. From this we have six inequalities:

i) 0 < k < 120 (first and second shows) ii) 600 < 6k < 720 (first and seventh shows) iii) 660 < 7k < 780 (first and eighth shows) iv) 540 < 7k < 660 (second and seventh shows) v) 660 < 7k < 720 (second and eighth shows) vi) 0 < k < 120 (seventh and eighth shows)

From these, we can get a combined inequality that is true for k = 109, llOor 111.

So, the problem has multiple answers. Thus, depending on the value of k, the first show can start any time between 12.00 hours and 12.09 hours and the duration of a show can vary between I hour 49 minute and 1 hour 51 minute.

Your Number Please

Yes she is right. In fact, my friend Dr R. Raghavendran of Kolkata has a proof of the same.

Let N = anan ,an 2 ... a2a,a() be the decimal representation of N.

Here, anan ,an 2 ... a2a,a0stands for

a„10n + a„. i 1° n" 1 + a n - 2 1 0 n 2 + •••+a2102 + a,101 +a„; where are integers with 0 < < 9, for i = 0, 1,2 n - 2, n - 1 and l < a n < 9 .


We know, as 99 is the LCM of 9 and 11, we see that N is divisible by 99 if, and only if, N is divisible by both 9 and 11. Now,

N is divisible by 9 <=> (an + an , + an 2 + ... + ^ + a, + a^ is divisible by 9 <=> the sum of the digits of N* (i.e. a„ + a, +%+ ... +an 2 + an , + an) is divisible by 9 N* is divisible by 9.

N is divisible by 11 {(-1 )nan + (-1 )n• 'an., + (-1 ) n • : \ _ 2 +...+ (-1 ) \ + (-l)'a, +a0) is divisible by 11.

The corresponding expression for N* is: {(-l)"a0 + ( - l ) n I a I + ( - l ) " - 2 a 2 + . . . + ( - l ) 2 a n _ 2 + ( - l ) i a n . 1 + a n } = ( - l ) " {a0 + ( - l ) 1 a 1 + ( - l ) 2 a 2 + . . . + ( - l ) n - 2 a n . 2 + ( . l ) n l a n . 1 + ( - l ) n a I 1 }

The obtained expression is divisible by 11 implies, N* is divisible by 11. Combining these bits of information we get that, N divisible

by 11 <=> N* divisible by 11. And we may conclude that, N divisible 99 <=> N* divisible by 99. So, Madame Serene is right.

Not a Game of Chess

Here is a solution from G. Ravi Chandra of Paloncha, Andhra Pradesh. Let us rewrite the problem as:

W X 8 Y Z A B j C D E F G H I

J K L M N O P

Q R S T U V

1

Since AB, when multiplied by 8. gives a two-digit number, it must be less than equal 12. Again, AB, when multiplied by W, gives a three-digit number JKL; so we get AB = 12, and W = 9. This gives us X= 0.

We also see, CDE = JKL = 108. We know, OP = 96. and Q = 1;


therefore, FG = MN = 97. Now, TUV = 108, therefore, QRS = 109, Y = 0 and Z = 9.

So, the division will look like,

90809 12)1089709

108 97 96

109 108

1

A Walk in the Garden

The minimum number of walks is five. The set of five instructions will be: 1) Left-Right-Left (LLR) 2) Left-Right-Right (LRR) 3) Right-Right-Right (RRR) 4) Right-Left-Left (RLL), and 5) Right-Left-Right (RLR)

Now, let XY denote the path 1 am standing on. I'm at the mid-point of XY, facing junction Y. Then, my answers after obeying the above instructions would be as given in the following table:

XY LLR LRR RRR RLL RLR

CA G G G F G

AC F G G G G

AB G G G F F

BA F F G G G

BC G G G G F


CB G F G G G CD G F G F G DC F G G G F AD G F F G G DA G G F G F BD G G F F G DB F G F G G

(Here, G = Grass and F = Flower.)

Since the combinations of all five answers are unique for each starting point and direction, it will be the minimum number of walks that I have to take for my daughter to ascertain my direction and starting point.

Marble Story

The total number of marbles is 2592. (Hint: Let the number be ABCD. Given, ABCD =A B xC D etc.)

Wordwallah's 'Words Mall'

Woody Wordwallah earned twenty-two silver coins on his ninety-ninth birthday.

Here Comes the Hangman

Let us define the three events.

A: X is hanged B: Y is hanged C: Z is hanged We note that originally the odds of X being released are simply 1/3. When the guard tells him that Y is to be hanged, his knowledge somewhat changes. So now, to determine the chances of X being released, in statistical terms, we look at the probability that 'X is released given that Y is hanged'.


If A (comp) is the complement of event A, meaning X is released, then we actually look at in statistical notation:

P (A (comp)/B} = P (A)*P (B/A), divided by [P (A)*P (B/A) + P( A(comp)*P(B/A(comp)}]

This is nothing but Bayes rule, and the result is 2/3. This amounts to saying that with the knowledge that Y is about to be hanged, the odds favouring X's release increase. We work on the assumption that the guard says the truth.

Ball Game, Sen Style

The answer is that the probability of drawing the red ball is the same anywhere you stand in the line. In picking up the red ball at any position, we have to consider not only the chances of picking the ball at that position but also the chances of all the boys in preceding places not drawing the red one.

Let Ai be the probability of drawing the red ball at the ith position, where i= 1,2 ... 32,33

Hence,

P (Al) = 1/33 P(A2)= P(A2n Al) + P(A2n Alcomp)

= 0 + P (A2/Alcomp)*P (Alcomp) = 1/32*32/33 = 1/33

This continues for all the positions in the line from 1 to 33 and the chances of picking the red ball are the same everywhere.

One for the Dog

The least number of biscuits is 3121. Let N be the number of biscuits and A be the number of friends then,

N = AA - A + 1 will give us the result. How? Well, that's another story.


Calculate, Do Not Guess

5929 = 529 x 11 + 110

Burrowing for Gold

This is a solution Swapan Kumar Chaudhuri, WBSEB, had sent. My doppelganger walks 30,000 m.

A X O

OABC is my tetrahedral dreamland with, AB = BC = AC = OA = OB = OC = 30,000 m

Suppose, I start from X, mid-point of AO, and burrow in a straight l-ne through the tetrahedron and emerge at Y, mid-point of BC My doppelganger also starts at X and takes the shortest route on the surface of the tetrahedron to reach Y.

A X O


If, the entire route from X to Y can be imagined on a plane, the shortest route is the straight line joining X and Y. To find XY. which lies on the surface of my dreamland, I place A OAC and A OBC on a plane having OB as a common arm.

Here, AB 11 OC, AO 11 BC and AB = OC, AO = BC or, OX = YC.

So, we have OX 11 YC. Therefore, OXYC is a parallelogram.

In OXYC, OC = XY = 30,000 m.

So, the shortest distance from X to Y over the surface of my dreamworld is 30,000 m!

A Quick Game with Your Friends

It is Jaypee. Such a sum is always divisible by 11. Here is an explanation. Suppose we take a number ABCD. The number can be represented as:

ABCD = 1000A + 100B + 10C + D

Now, we take the first digit to the end and get the number BCDA,

Or, DABC = 1000D + 100A + 10B + C

Now, ABCD + DABC = (1000A + 100B + 10C + D) + (1000D+ 100A+ 10B + C)

= 1100A+ 110B+ 11C+ 1001D = 11 (100A+ 10B + C + 91D)

So, the sum is always divisible by 11.

Who Scored More? Both got equal marks. Here is a solution from Satya Kiran Thapa:

The problem can be represented as 7* = Ky + R, where x e N, K = 11 or 13, y is the quotient and, obviously, 0 < R < K. Here N is the set of integers.

If we want to increase the power of 7 by one, we will have: 7"+ 1 = 7Ky + 7R.


It may so happen that 7R > K; in that case, 7R may be represented as: 7R = Kz + R*, where z is the quotient and R* the new remainder. Again 0 < R* < K. Then, we get,

7X +1 = (7y + z) K + R*. Using the above information we get,

7711, when x = Remainder 1 7 2 5 3 2 4 3 5 10 6 4 7 6 8 9 9 8 10 1

We observe that the remainders obtained from 7V11 follow a pattern as x increases from 1 to 10 and the same pattern is repeated in batches of 10. Or, remainders of 7(x + I0>"/11 are equal for a particular x and varying y.

The same can be applied when T is divided by 13. In this case, the pattern emerges after 12 and is repeated in batches of 12. Or, remainders of 7,x + l2y)/13 are equal for a particular x and varying y. The following table gives the first 12 remainders.

7713, when x = Remainder 1 7 2 10 3 5 4 9


5 11 6 12 7 6 8 3 9 8 10 4 11 2 12 1

Now, 123456789 = 12345678 x 10 + 9.

So, when 7123456789 is divided by 11, the remainder will be equal to the remainder when 79 is divided by 11. From the first table, we find that the remainder is 8. Again, 123456789 = 10288065 x 12 + 9.

So, when 7123456789 is divided by 13, the remainder will be equal to the remainder when 79 is divided by 13. From the second table, we find that the remainder is 8.

PM Wins Thirteenth Round

Pollster Manipulator got 2470 votes. (Hint: Add I2 + 22 + 32 + 42 + ... + 172 + 182 + 192)

Santa's Order

From front to back: Prancer, Cupid, Rudolph, Dasher, Blitzen, Vixen, Comet, Donder, then Dancer.

Trick Unmasked

l et b(n) be the mean number of birthdays for n people. Then b(n) = b(n - 1) + 1 - b(n - l)/365


b(n) = 1 + 364/365*b(n - 1) b(n) = 1 + 364/365 + (364/365)2 + ... + (364/365)<n" » b(n) = (1 - (364/365)"}/(l - 364/365) = 365{1 - (364/365)"} =

365 - {364"/365<n" "}

The number of days worked is 365 - b(n) = 3647365<"- '>

And the total number of worker-days is n*364"/365<n' "

Differentiating and setting to zero, the maximum is at 364.5. In fact, the number of worker-days is the same at 364 and 365. So 364 workers should be hired and the last salary saved. The expected number of worker-days is 48943.5.

Thrice as Big

Since the first digit of the number is 1, the number can be written as 10m + x, where x is an m-digit number. Transferring the 1 to the end of the number gives us the number lOx + 1. We must therefore solve the equation lOx + 1 = 3(10"1 + x), or equivalently 7x + 1 = 3*10m. In other words, the remainder upon dividing 3*10™ by 7 must be 1. We can either find the smallest m by trial and error or perform the following long division until a remainder of 1 is obtained:

42857 7)3000000...

28 20 14

60 56

40 35

50 49

1


So x = 42857 does the trick and the number we seek is 142857.

lake Your Pencil for a Walk

Here is one solution and there are 11 more possible!

The Challenge

Here is why:

M2 - N2 =A3 or, (M - N)(M + Y) = A3.

One possible solution can be derived from

M - N = A, M + N = A2

Which implies M = A(A + l)/2, N = A(A -l)/2.

M, N are clearly integers as A(A + 1) and A(A -1) are always even.

The Second Visit

The answer is 17 days and 3 hours later, which would have been a Wednesday. This is the only other time in the same month when the two would agree at all.

In 17 days, the slow clock loses 17*24*7 minutes = 2856 minutes, or 47 hours and 36 minutes. In 3 hours more, it loses 21 minutes, so it has lost a total of 47 hours and 57 minutes. Modulo 12 hours, it has 'gained' 3 minutes, so as to make up the 3 minutes it was slow on Sunday. It is now (fortnight plus 3 days) exactly accurate.


Since the clock was not adjusted since the last visit, it's also possible that the radio time shifted by one hour due to a change to or from summer daylight saving time. However, it turns out that the only additional possibilities that need to be considered are those of 4 days 15 hours later, when the clock would have lost 12 hours 57 minutes, and 29 days 15 hours later, when the clock would have lost 3 days 10 hours 57 minutes. Without even considering the rules for when in the month the clock is changed, these possible solutions are ruled out because we know that both visits were in the evening.

Know Your Dish

Let the dishes be A, B, C, D, E, F, G, H and I. At the first meal, order dishes A, B, B, C and D. At the second meal, order dishes D, E, E, F, and G. At the third meal, order dishes G, H, H, I, A.

Game Show Winners

Since no family can get exactly four correct matches (because four correct implies getting all five correct), and since all the scores are different, the scores must be 5,3,2,1, and 0. In succession, assume each family has a perfect score, and determine the scores of the other families. Only when the Flintstones have a perfect score are all the other scores different.

Get Rich Quick

Stephen can take 90 gems safely. He can take all gems except those whose numbers are perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, and 100).

Getting Better on the ELO Ratings

From B's statement we conclude that A had at most three wins, giving him a maximum of 3.0 points. This leaves at most 2.5, 2.0, 1.5 and


1.0 for the other four players. These five results sum up to 10.0 points, exactly the sum of points possible. Thus, we know the score for each of the contestants.

For A to get 3.0 points he must have won against C, D and E (and lost to B). B reached 2.5 points by tying against C, D and E (and winning against A).

Looking only at C, D and E, we find that C scored 1.5, D scored 1.0 and E scored 0.5. From E's statement we find that D must have scored a win and a loss. C obviously scored a win and a draw and E scored just one draw and a loss.

Now the score table is complete:

Players A B C D E Total Points

A X 0.0 1.0 1.0 1.0 3.0

B 1.0 X 0.5 0.5 0.5 2.5

C 0.0 0.5 X 1.0 0.5 2.0

D 0.0 0.5 0.0 X 1.0 1.5

E 0.0 0.5 0.5 0.0 X 1.0

Fundoo Island

The maximum number of roads is 12. The maximal number of roads on an island with n towns is 3*(n - 2) for (n > 2). This result is common knowledge in graph theory: A planar graph with n nodes has at most (n - 2)*3 edges (n > 2).

Jousting Knights

When the knights meet the first time, they together had covered the length of the arena once. When they meet again, they together rode three times the length of the arena. So each of them had travelled three times as far as when they had met first. Looking at the Red Knight, this leads to the following equation:


Arena + 50m = 3*60m

Hence, the arena is 130m long.

Postage Stamp Geometry

Fold the stamps along the line between 1 8 2 7 and 4 5 3 6. Push the 7 and 6 pair between the 1 and 4 pair. Now push the 7 and 6 pair further in between the 8 and 5 pair, tucking the 4 between the 3 and the 5.

Puzzlemania II

Puzzledom Open

Puzzledom Open, the most prestigious lawn tennis tournament in Puzzlenia, was being played. Pace had defeated Grace in the first semi-final, while Ace had felled Menace in the second. Tom had missed both the semi-final matches, and was dying to know the scores. His luckier buddy Dick gave him cryptic details: 'Something strange happened in both the matches. In the first match, after the Nth set, the total number of games played up to that point was prime when N was odd, and a perfect square when N was even. In the second match, exactly the opposite thing happened. That is, after the Nth set, the total number of games played up to that point was prime when N was even, and a perfect square when N was odd. Now that's what you call a coincidence!'

TOM: And I missed such interesting matches. (Heaves a sigh) But, tell me the scores.

DICK: The difference between the total number of games won by the winner and that won by the loser in both the matches were the same, and it's that number over there. (Points to a number in the calendar.)

Tom was then able to figure out the scores. Can you?


Same Old Boat Ride (But With Cannibals)

Four missionaries and four cannibals are standing on the bank of a river, and want to cross over to the other side. The only means to cross the river is a single boat that can hold at most four persons in one trip. A person has to pay for the boat ticket for each one-way trip he undertakes. Now, the travellers want to get to the other bank as quickly and as economically as possible. Also, both the missionaries and the cannibals wish to incur the same expenditure in order to cross the river.

However, if the cannibals ever outnumber the missionaries on either bank or on the boat, the missionaries will be eaten.

How should they cross the river?

Integer Dialogue

This is a variant of a famous puzzle. Two different integers, greater than 1 and less than 100 are chosen. Now, Tom is told the sum of the two integers, while Dick their product.

a) Based on the following conversation between Tom and Dick, can you determine the integers?

TOM: I do not know. D I C K : 1 do not know either. TOM: I still do not know. D I C K : I still do not know. TOM: I can't tell. D I C K : I can't tell either. TOM: Yes, got the numbers. D I C K : M e t o o .

b) What if the conversation proceeded as?

D I C K : I do not know.

TOM: I do not know either. D I C K : I still do not know. TOM: I still do not know. D I C K : I can't tell. TOM: I can't tell either. D I C K : Yes, got the numbers. T O M : Me too.

Name Game 1

P u z z l e m a n i a I I 159

The following are the abbreviated titles of books, containing numbers. Next to each title, the author's initials have been included in brackets. For example: The 3 M (A.D.) means The Three Musketeers (Alexander Dumas).

Can you decipher the titles and the authors of the following list? (In some cases, I have given two options separated by /. You can choose one or both.)

1) C t o O ( K . F ) 2) The P of 1 (B.C.) 3) A T of 2 C (C.D.)/2 R M in C (T.L.) 4) 3 M in a B, and 3 M on the B (J.K.J.) 5) 4 B M (J.P.) 6) 5 Q of the O (J.H.)/L at 5 (R.B.) 7) P 1919: 6 M that C the W (M.O.M, R.H.) 8) The 7 S (M.D.)/7 D S (M.B.) 9) H 8 (J.E.)/8 S G (B.H.)

10) 9: A N of S (J.B.)/C 9 (L.R.) 11) 10 L I (A.C.) 12) 11 D: A N of the H (D.H.) 13) 12 R H (J.A.) 14) 13 D (A.C.)ZThe 1 3 G S ( P O ' B )


15) 14 : G U A in a C (S.Z.) 16) The 15 S (C.C.) 17) The 16 P: A N (R.H.)

Name Game 2

The following are the abbreviated titles of films, containing numbers. For example: The 3 M means The Three Musketeers.

Can you decipher the titles of the following list? (In some cases, I have given two options separated by /. You can choose one or both.)

1) OE/B 2 0 2) 1 F D/A F 1 3) The M H 2 F 4 ) . 3 D of the C 5) S H—The S of 4/The M of 4 S 6) The 5 H/5 E P

7) 6 D of S 8) 7 Y in T/The M 7 9) 8 M O

10) The W 9 Y 11) The 1 0 C / 1 0 T I H A Y 12) O i l 13) 12 0 C H / B the 12-M R 14) 13 C A I T 15) 14 H/14 D in M

Travel Chat

While on a globetrotting spree, Tom and Dick kept in touch with each other via the Internet, mostly through chat rooms. Whenever


a user logged out of a chat room, the chat applet showed the logout time, like your computer clock, to the other person.

Here's a sample of their chat conversation:

TOM: Today I saw a statue with the head of a lion and the body of a fish. DICK: Hey, I was just visiting a structure with the head of a man and the body of a lion. At the end of the chat, Tom logged out, and Dick immediately followed suit. They chatted again some time later, and their logout times were exactly the reverse of their earlier logout times. In other words, Tom's logout time was the same as Dick's earlier logout time. And Dick's logout time was the same as Tom's earlier time. Quite a coincidence!

Can you guess where Tom and Dick are?

King-Queen-Jack

Place a Jack (J), a Queen (Q) and a King (K) on three slots in a row, say, in the order JQK (J on extreme left, K on extreme right). Now, reverse the position of the cards (i.e. KQJ) in the least possible number of moves. In a valid move, a card can be moved either left or right into an empty slot or placed onto a card of a higher rank. For example, a J can be placed on a K, but not vice-versa. Also, only the top card of the stack can be moved.

What is the least number of moves required if:

a) There are only three slots? b) There is an empty slot to the left of the Jack (initial

arrangement is JQK)? c) There is an empty slot to the right of the King (Initial

arrangement is JQK)?


Triangular Number Heap

Tom writes some numbers in a triangular form as:

1

2 3 4 5 6 7 8 9 10 11 12 13 14 15

His buddy, Dick, writes the same numbers in a triangular form in the following manner:

1

2 6 3 7 10 4 8 11 13

9 12 14 15

In both cases, the number of rows (N) = 5 You'll notice that numbers 1, 2, 8, 14 and 15 at both times

occupy identical positions. For what values of N would the instances of such numbers be odd?

Knotting a Tie

There are many ways to knot a tie, but the Pratt knot (more famous as the Shelby knot) has a peculiar characteristic, distinct from the other methods. The tie starts 'inside out', i.e. the seam of the tie faces outwards. Here's how a Shelby knot can be tied:

1) Start with the tie inside out, the wide end A under the narrow end B. (Assume that the tip of A is to the right of the tip of B.)

2) Take A over B and loop it down between the collar and the tie.

P u z z l e m a n i a I I 177

3) Take A over to the left. 4) Pull A up, behind the loop. 5) Bring A down through the knot and tighten.

Here's another very simple method:

1) Start with the tie inside out, the wide end A under the narrow endB.

2) Take A over and across to your left. 3) Pull A up, behind the loop. 4) Bring A down through the knot and tighten.

Can you find other ways to knot a tie, starting with the tie inside out? All the methods I have found start with the wide end under the narrow end. Is it possible to knot a tie, starting inside out, with the wide end over the narrow end?

Number Search

Harry gave a slip of paper each to Tom and Dick. Tom's paper had the last digit of the sum of the cube and the square of an integer (whose last digit is non-zero), while Dick's paper had the last digit of the difference of the cube and the square of that same integer.

H A R R Y : Tom, do you know Dick's number? T O M : N O .

H A R R Y : Dick, do you know Tom's number? D I C K : N O .

H A R R Y : Tom, do you know the last digit of the integer? T O M : Yes. Can you determine the last digit of the integer? What if Harry had received the same sequence of answers by asking Dick first, then Tom and finally Dick again?


Age-Old Tale

Al and his sons Bob and Chris had gathered to celebrate grandpa Don's birthday. At the dinner table, they started discussing their ages, and discovered some amazing facts:

1) The last digit of Chris' age was the first digit of Al's age. 2) The last digit of Bob's age was the first digit of Don's age. 3) The first digit of Chris' age and the last digit of Don's age

were the first and the last digits of the sum of the ages of three of them.

4) The first digit of Bob's age and the last digit of Al's age were the first and the last digits of the sum of the ages of all four.

5) Both sums (in 3 and 4) had the same middle digit.

If all the ages were below 100 and above 10 years, determine their ages.

Chess Classic 1

Place one queen, three kings, two rooks and one knight on a standard 8 x 8 chessboard so that every square (occupied or not) is under attack by at least one chess piece.

Chess Classic 2

Place one queen, three kings and three rooks on a standard 8 x 8 chessboard so that every square (occupied or not) is under attack, and the queen cannot attack any of the three kings.

Palindromes

Find the smallest multi-digit perfect square, which is a palindrome—a number that reads the same from both ends, for example 131—all of whose digits are a) even, and b) odd.


Numbers on a String

Suppose an integer N (in base 10) requires M digits to represent it in base B (B > 1). For example, the number 12 (in base 10) is represented as 1100 in base 2, so that M = 4.

Now, create all the substrings of this representation in base B. In our example, the substrings would be:

Substrings of length 1 = 1, 1, 0, 0 Substrings of length 2 = 11, 10, 00 Substrings of length 3 = 110, 100 Substrings of length 4 = 1100

Now, list all the 'distinct' substrings (without leading 0s). They are: 0, 1, 10, 11, 100, 110, and 1100.

Let T represent the total number of distinct substrings of N (which requires M digits to represent it in base B). In our example, T = 7.

What is the minimum value of T (in terms of N, M and B)?

Pirates Ahoy!

A handful of pirates distributed some golden coins among themselves, such that:

The share of Pirate 1 plus half the total share of the rest (all pirates except #1) = the share of Pirate 2 plus 1/3 of the share of the rest (all pirates except #2) = ... = the share of Pirate N plus 1/(N + 1) of the share of the rest (all pirates except #N). Captain Cutthroat, the puzzle-loving pirate, informed his captive Jolly Rogers of this strange method of distribution. Moreover, he promised to set Jolly free if he could correctly tell him the share of each pirate.

'The total number of coins is less than 1000, and more than 50% of us received an odd number of coins.'


Jolly Rogers got to work immediately, and after some time, asked the Captain, 'Can you give me a little hint?'

'The least a pirate got is this number.' Jolly Rogers was still not sure until the Captain boasted,

'And I got more than 10 times that fellow.' Can you guess the number of pirates and the share of each?

Polygon Trap

In a 2N-sided regular polygon, join the mid-point of each side with the end points of the opposite parallel side. The lines crossing over inside the polygon enclose a new polygon.

Find the ratio of the areas of the new and the original polygon.

Zero? That's Right!

A teacher asked his students to take the digits 0 through 9, and arrange them in two numbers. He further asked them to find the least possible difference between these numbers. One of the students declared that the least possible difference is 0. Can you explain how?

Code Cracker

I have written a code in which each of the digits 0 through 9 represents some other digit. All perfect square integers below 100 in my code, when decoded, result in primes. Exactly half of all the perfect square integers below 1000 in my code, when decoded, result in primes. What is my code? In other words, what does 0123456789 in my code actually represent?

(The source of all the puzzles above, and their solutions, is Sudipta Das, Kolkata.)


Russian Doll

Inside a box, there are twelve hollow Russian dolls of different sizes. The smaller dolls can fit inside one another, giving one outer doll and eleven inside. A person sits down with this box. He picks up two dolls at random, compares their sizes by trying to make one doll fit inside the other and puts the smaller doll inside. Then he picks up a third doll and starts comparing with the outermost doll and fits it. This way, he continues picking up, at random, dolls from the box and fitting them after comparing. How many comparisons will he have to make to fit all of them into one? (Amitabh Roy Sharma)

Got Zapped!

My young Number Tumbler Club members have recently started exploring the English alphabet. They chanced up on a book by my guru Martin Gardner and found a way of doing alphabet arithmetic. Last meeting was quite stormy; they came up with many examples and I rejected them. Now, my afternoon naps are interrupted by phone calls from these members. I try and mumble some answers but most of the time I 'm way off the mark. Here is one such example:

Look closely at the equation, ZZ + AA + PP = ZAP. You are told, each given letter stands consistently for the

same digit and different letters stand for different digits throughout this equation. What is ZAP in digits?

Useless Cricket Statistics

The rain gods were very angry on the fourth day of the test match. The umpires decided to call off the day. With very little


to do, Tendulkar, Ganguly and Kumble decided to play around with cricket statistics. They were trying to find sets of three two-digit positive integers (runs scored by Indians in the first innings of the match) such that if they added any two integers of the set together they formed a perfect square.

Kumble decided to add more to the game: Within each set, the three integers have to be different, but an integer used in one set could be used again in another. They found two such sets, such that the six perfect squares they each formed from their two sets were all different. One of Kumble's sets was the same as one of Tendulkar's and the leg spinner's other set was the same as Ganguly's; Ganguly's and Tendulkar's other sets were different.

Now, tell us: 1) What were the integers in each of Kumble's two sets? 2) Which integer appears in two sets that none of them found?

Follow Me

Instructions

In this maze, follow the edges of the triangles. You can go through a circle, but you must alternate going through black, white, black, white, black, white, etc. You cannot reverse your direction at a


circle; you must keep going through it. Find the shortest path through the maze. (Jim Loy)

Playing With Symbols Really

All you have to do is to find the missing term, denoted by the question mark, in the following series:

I, III, 110,10, ? , I0I0I

Words, Words I

What is common to these words?

R A N D O M , TANDEM, G A N D E R , HANDLE, SANDY, PANDEMONIUM

Words, Words II

A G O N I Z E , A G R A R I A N , A G R O U N D , A G R I M O N Y , AGREEABLE and AGITPROP

Which of the above words is the next in the following series?

AARDVARK, ABDUCTOR, ACADEMIA, ADULTERY, AEROBICS, AFFINITY, ?

Words, Words III

Find all the words that can appear as the next word in the following series:

GENTOO TOOLKIT KITTEN


TENDER DERIDE ???

Carroll Twist

A set of puzzles without Lewis Carroll's doublets is difficult to imagine. All one needs is to start with a word, change one letter at a time, creating meaningful words and reach a 'destination' word. In this case, all you have to do is turn BEAR into CROW. No proper nouns please.

Sentence Walk

Look at the grid closely. There are letters and blank squares. All you have to do is create a meaningful sentence by starting from E in row 1, column 1 and ending in D in row 6, column 1.

Here are some rules:

1) You have to go through every square. You move from one square to the other in any direction without skipping.

E V N H D

E N O E A R

A S I H

N E I N T

I E D N

N S T F O U

Figure 17


2) Words are made by moving from one letter to the other, and after a word is completed, one moves to a blank square before starting another word.

3) All words, including proper nouns, are admissible as long as the final sentence is meaningful.

Valid Time Instant

At a certain time instant, the minute hand M and the hour hand H of a watch are in such a configuration (C) that if their positions are exactly reversed then the resultant position also corresponds to a 'valid time instant'. A valid time instant means that under normal circumstances M and H can have that configuration at some time or the other. The question is: How many such Cs are possible? (Arun Kumar Saha)

The House of Mr X

Suppose all the houses on a road are on one side of it. The houses are numbered consecutively. The sum of the numbers on the right of Mr X's house is equal to that of the houses on the left. If there are more than 50 and less than 300 houses on the road, what is the number of Mr X's house?

Note, folklore has, years ago when this puzzle appeared in a magazine, Srinivasa Ramanujan found the general solution of all such problems in a few seconds. (Pritibhushan Sinha)

Gates Got It!

Internet buzz says, an IT tycoon asked Bill Gates this. Mr Microsoft could not be misled, though.

There are four dice. Die A shows the numbers: 1, 2 ,3 ,9 , 10


and 11 on its six faces; Die B shows the numbers: 0, 1, 7, 8, 8 and 9 on its six faces; Die C shows the numbers: 5, 5, 6, 6, 7 and 7 on its six faces; and Die D shows the numbers: 3 , 4 , 4 , 5 , 11 and 12 on its six faces. One has to choose a die such that his chances will be higher to roll a number larger than that of the die chosen by another person after him but rolled simultaneously. What should be one's choice?

Stack It Up

A total of n cards, numbered 1 through n, are divided into two stacks. What is the minimum value of n such that at least one stack will include a pair of cards whose numbers add up to an exact square?

Liar Playing Spitball

It was the last class of the afternoon and Mrs Nesbitt was reading from a book as her seventh grade algebra class finished its homework. Being engrossed in her reading, she had her nose buried in the book. All of a sudden, from the right-hand side of the room, a big, well-constructed spitball sailed past her head and slapped against the blackboard behind her. Glancing up startled, she could only tell that it had come from the row closest to the window or the row immediately inside of that. Also from the angle, it appeared to come from the front half of one of those rows. All the students were deeply engrossed in their work and no one was looking up from that side of the room. From the other side of the room, however, it was a different story. Several students were looking around with a grin on their faces as if they had seen the whole thing. But of course, no one would


admit to anything. So Mrs Nesbitt was left to discover the culprit herself. She narrowed the field of possible delinquents to four students and began to question them. She asked, 'Who threw that spitball?' Each of the four students made one statement.

M A R Y A B R A M S O N : I didn't do it. J A C K S O N : I didn't do it. M A R K M A R T I N : Abramson did it. Lu ANN: Mark is telling the truth.

Being a very clever teacher. Mrs Nesbitt knew that only two students were telling the truth. So she questioned them again. She asked again, 'Who threw that spitball?'Again each student made one statement.

M A R Y : L U d i d i t .

ANN: Mary is lying. M A R T I N : Ann didn't do it. J A C K S O N : Martin didn't do it.

'Now this is interesting,' thought Mrs Nesbitt, 'Each student who lied earlier has told the truth, and each one who told the truth has now lied! Oh well, no matter, I know who the culprit is!'

Do you? (Sally Quinn)

Lunchtime Variety

Grant, Neal and Kyle just finished their lunch. One of the boys had a ham sandwich, another a turkey sandwich and the third, a chicken sandwich. If only one of the following four statements is true, who had what for lunch?

a) Kyle had the turkey. b) Kyle did not have the chicken.


c) Neal did not have the chicken. d) Neal did not have the ham.

On Track

Every station on the Eastern Railway sells tickets to every other station. When new stations were added, 46 additional sets of tickets had to be printed. Tell me, how many new stations were added, and how many stations were there before the additional stations came up?

Get the Arrangement

See the arrangement below and place a digit from 0 to 9 into each of the seven locations, such that each of the three numbers read from top-to-bottom in each of the three directions is divisible by 7. You can use a single digit more than once, but each of the nine multi-digit numbers must be unique. (The numbers are the two-digit numbers AC, AB, BE, CF, EG and FG, and the three-digit numbers ADG, BDF and CDE.) To minimize variations, choose B < C, and E < F. (K.D.)

A B C

D E F

G

On Its Head

The two-digit number 69 reads the same when read upside down. Which two-digit number increases by the least non-zero amount when read so?


A Test for Sanity?

Four mathematicians make the following statements:

A L I C E : I am insane. BOB: I am pure. CHARLIE: I am applied. DOROTHY: I am sane. A L I C E : Charlie is pure. BOB: Dorothy is insane. CHARLIE: Bob is applied. D O R O T H Y : Charlie is sane.

Describe the four mathematicians, given that: • Pure mathematicians tell the truth about their beliefs. • Applied mathematicians lie about their beliefs. • Sane mathematicians' beliefs are correct. • Insane mathematicians' beliefs are incorrect. (Mathematics and Informatics Quarterly)

Love = A White Mouse

Fluffytail the cat finds himself encircled by thirteen mice. There are twelve grey mice and one white mouse. Fluffytail is on red alert. Suddenly, he sees his friend Yellowpatch winking at him. 'FT,' she says, 'You catch every thirteenth mouse going in one direction only round the circle. And the last mouse you catch for me must be white. Then you can win my love.' Fluffytail now has a problem, with which mouse should he start? (B.K.)

Date of Birth

During New Year's Day in the year 2002, when Mr Girard queried Mr de Marco about his date of birth, Mr de Marco said, 'I was


born on the pth day of the qth month in a certain year in the last century where both p and q are factors of the numerical magnitude of the current year (2002). This year, my birthday occurs on a Thursday, when I will be exactly (pq)2 years old.' Mr Girard requested for some additional information, when Mr de Marco replied, 'Neither p nor q are equal to 1. Moreover, (p + q) is expressible as the square of an even number.' Determine the precise date of birth of Mr de Marco from the information provided in the above statements. (K. Sengupta)

Time Again

If 'SO' corresponds to 17:23 hours and 'MY' corresponds to 01:13 hours, deduce the precise time corresponding to 'AT' in a twenty-four-hour watch. (K. Sengupta)

Solutions

Puzzledom Open

First semi-final: 6-1, 6-3, 6-1

Second semi-final: 6-3,4-6, 6-0, 6-0

Standard tennis rules state that a set can be won in 6,7,8,9,10,12 or 13 games. These correspond to the scores 6-0, 6-1, 6-2, 6-3, 6-4, 7-5 and 7-6. Also, in some tournaments, the final set has to be won by at least two clear games. This is the case in Wimbledon, and not in the US Open. But it does not affect the final solution in any way.

List out all possible scores for both the semi-finals, and compute the difference for each. Since Tom was able to figure out the scores just by looking at the difference, this difference must be unique for each list.

A difference of 13 is unique for each list of cases.


Same Old Boat Ride (But With Cannibals)

Two possible ways to cross the river:

(M = Missionary and C = Cannibal)

Option A

Origin Boat Ride Direction Destination

4M,4C (initial condition) 3M, 3C — > 1M, 1C

4M, 3C <r~ 1C

3C —> 4M, 1C

4C <— 4M 4M,4C

Option B

Origin Boat Ride Direction Destination

4M,4C (initial condition) 4M 4C

4M, 1C <— 3C

1C — » 4M, 3C

1M, 1C <r- 3M, 3C 4M.4C

Total number of trips = 5 (as quickly as possible)

Total number of tickets bought = 12 (as economically as possible)

Also, the missionaries and the cannibals incurred the same expenditure.


Integer Dialogue

A) The numbers are: 70 and 99. Tom got 169. Dick got 6930. And the solution is unique!

B) The numbers are: 77 and 90. Tom got 167. Dick got 6930. And again, the solution is unique.

Name Game 1

1) Code to Zero (Ken Follett) 2) The Power of One (Bruce Courtenay) 3) A Tale of Two Cities (Charles Dickens)/7Wo Rivers Meet in

Concord (Thomas Longstreth) 4) Three Men in a Boat, and Three Men on the Bummel (Jerome K.

Jerome) 5) Four Blind Mice (James Patterson) 6) Five Quarters of the Orange (Joanne Harris)/L/ve at Five

(Raymond Benson) 7) Paris 1919: Six Months That Changed the World (Margaret

Olwen MacMillan, Richard Holbrooke) 8) The Seven Sisters (Margaret Drabble)/SeveM Deadly Sins

(Michael Bishop) 9) Hard Eight (Janet Evanovich)/£;£/» Skilled Gentlemen (Barry

Hughart) 10) Nine: A Novel of Suspense (Jan Burke)/Cloud Nine (Luanne

Rice) 11) Ten Little Indians (Agatha Christie) 12) Eleven Days: A Novel of the Heartland (Donald Harstad) 13) Twelve Red Herrings (Jeffrey Archer) 14) Thirteen Diamonds (Alan Cook)/The Thirteen Gun Salute

(Patrick O' Brian) 15) Fourteen: Gmwing Up Alone in a Crowd (Stephen Zanichkowsky) 16) The Fifteen Streets (Catherine Cookson) 17) The Sixteen Pleasures: A Novel (Robert Hellenga)


Name Game 2

1) Zero Effect/Bravo Two Zero 2) One Fine Day/Air Force One 3) The Mirror Has Two Faces 4) Three Days of the Condor 5) Sherlock Holmes—The Sign of Four/The Merchant of Four

Seasons 6) The Five Heartbeats!Five Easy Pieces 7) Six Degrees of Separation 8) Seven Years in Tibet/The Magnificent Seven 9) Eight Men Out

10) The Whole Nine Yards 11) The Ten Commandments! 10 Things 1 Hate About You 12) Ocean's Eleven 13) Twelve O'Clock High/Beneath the Twelve-Mile Reef 14) Thirteen Conversations About One Thing 15) Fourteen Hours/Fourteen Days in May

Travel Chat

Tom is in Singapore (merlion—a creature with the head of a lion and the body of a fish; Singapur, Lion City).

Dick is in Cairo (visited the Sphinx of Giza).

Suppose Tom's initial logout time was Ht:Mt; and Dick's was Hd:Md. Then, their next logout times were:

Tom: Hd:Md Dick: Ht:Mt

If Ht > Hd, the time difference (in minutes) between the two places = 60*(Ht - Hd) + (Mt - Md)

Time difference (for first chat) = 60*(Ht - Hd) + (Mt - Md)


Time difference (for second chat) = 60*(Hd - Ht) + (Md - Mt) + 60* 12

So, 60*(Ht - Hd) + (Mt - Md) = 60*(Hd - Ht) + (Md - Mt) + 60*12 => 60*(Ht - Hd) + (Mt - Md) = 360

Time difference = 6 hours

If Ht < Hd, then the time difference (in minutes) between the two places = 60*(Ht - Hd) + (Mt - Md) + 60* 12

Time difference (for first chat) = 60*(Ht - Hd) + (Mt - Md) + 6 0 * 1 2

Time difference (for second chat) = 60*(Hd - Ht) + (Md - Mt)

So, 60*(Ht - Hd) + (Mt - Md) + 60* 12 = 60*(Hd - Ht) + (Md - Mt) => 60*(Ht - Hd) + (Mt - Md) + 60* 12 = 360

Time difference = 6 hours

In either case, the time difference between the two places is 6 hours. The clues about the cities may be enough to solve the puzzle.

The time difference also satisfies the cities:

Cairo (G.M.T. + 2:00) Singapore (G.M.T. + 8:00)

King-Queen-Jack

a) Unique solution is 20 moves: J: Right-Right, Q: Left, J: Left-Left, K: Left, J: Right-Right, Q: Right, J: Left-Left, Q: Right, J: Right-Right, K: Left, J: Left-Left, Q: Left, J: Right-Right.

b) Ten-move solution: J: Left, Q: Left, K: Left, J: Right-Right-Right, Q: Left, K: Left, Q: Right-Right.

c) Ten-move solution: J: Right-Right-Right, Q: Left, K: Left, Q: Right-Right, K: Left, Q: Left, J: Left.


Triangular Number Heap

In Tom's triangle, a number in the Pth row, Qth column has a value P(P -1)/2 + Q. In Dick's triangle, a number in the Pth row, Qth column has a value (Q/2)*(2*N - Q + 1) - (N - P).

We are looking for numbers in the Pth row, Qth column having the same value in both triangular forms. So, P(P - l)/2 + Q = (Q/2)*(2*N - Q + 1 ) - ( N - P )

=> P = [3 + - V{(2*Q - 3)2 + 8*(Q - 1)*(N - Q)}]/2

Now, P and Q must both be integers, with Q < P < N. The instances of such numbers occurring are odd for N = 1 (1), 2 (3), 5 (5), 22 (5), 121 (7), 698 (11), 4061 (5), 23662 (5)...

The number in brackets is instances of such numbers. The identical position numbers for N = 22 are 1,2, 127 (sixteenth row, seventh column), 252 and 253.

Knotting a Tie

Here are some methods, assuming that, at the start, the tip of A is to the right of the tip of B: Method 1 1) Start with the tie inside out, A under B. 2) Take A over and across to your left. 3) Take A underneath and back across to your right. 4) Lift A up and loop it down between the collar and the tie. 5) Pull A down to your left. 6) Take A over and across to your right. 7) Loop A under and up between the collar and the tie, and pull down. 8) Bring A down through the knot and tighten. Method 2 1) Start with the tie inside out, A under B. 2) Take A over B and loop it down between the collar and the tie. 3) Pull A to your left.


4) Lift A up, and loop it down between the collar and the tie. 5) Pull A down to your right. 6) Take A over and across to your left. 7) Loop A under and up between the collar and the tie, and pull down. 8) Bring A down through the knot and tighten.

Method 3 1) Start with the tie inside out, A under B. 2) Take A over B across to your left. 3) Loop A under and up between the collar and the tie, and pull down. 4) Pull A down to your right. 5) Take A under and to your left. 6) Lift A up, and loop it down between the collar and the tie. 7) Pull A down and to your right. 8) Take A over and across to your left. 9) Loop A under and up between the collar and the tie, and pull down.

10) Bring A down through the knot and tighten.

Method 4 1) Start with the tie inside out, A under B. 2) Take A over B and loop it down between the collar and the tie. 3) Pull A down to your left. 4) Lift A up, and loop it down between the collar and the tie. 5) Pull A down to your right. 6) Lift A up, and loop it down between the collar and the tie. 7) Pull A down to your left. 8) Take A over and across to your right. 9) Loop A under and up between the collar and the tie, and pull down.

10) Bring A down through the knot and tighten.

Number Search

The following table lists the possibilities:

Last digit of the integer: 1, 2, 3, 4, 5, 6, 7, 8, 9 Tom's number: 2, 2, 6, 0,0, 2, 2, 6, 0


Dick's number: 0,4, 8, 8,0,0, 4, 8, 8

If Tom's number = 6, then he could have deduced that Dick's number = 8. Dick concludes, from his negative reply, that his number is either 0 or 2.

If Dick's number = 4, then he could have deduced that Tom's number = 2. Also, if Dick's number = 8, then Tom's number has to be 0 (as Dick has already deduced that it can't be 6). His negative reply helps Tom to deduce that Dick's number is 0.

As Tom can deduce the last digit of the integer, his number must be 0, and the last digit of the integer must be 5.

Let's now see what happens if Harry receives the same sequence of answers by asking Dick first, then Tom and finally Dick again.

If Dick's number = 4, then he could have deduced that Tom's number = 2. Tom concludes, from his negative reply, that Dick's number is either 0 or 8.

If Tom's number = 6, then he could have deduced that Dick's number = 8. Also, if Tom's number = 2, then Dick's number has to be 0 (as Tom has already deduced that it can't be 4). His negative reply helps Dick to deduce that Tom's number is 0.

As Dick can deduce the last digit of the integer, his number must be 0, and the last digit of the integer must be 5.

Age-Old Tale

Suppose Chris' age was (10*a + b) years, Al's age was (10*b + c) years, Bob's age was (10*d + e) years and Don's age was (10*e + f) years. Let the common middle digit, as mentioned in condition 5, be g.

The sum of the ages of three of them < 99*3 = 297. So, Chris' age is < 29.

The sum of the ages of all four < 99*4 = 396. So, Bob's age is <39.

We can further conclude that the sum of the ages of all four < 99*2 + 29 + 39 = 266. So, 'd' is either 1 or 2.


Condition 5 implies that one of the four ages was 100*(d - a) + (c - 0 = A (say).

Now, only d = 2 and a = 1 ensures that 10 < A < 100. As f - c < 9, then A = 100 - (f - c) 91 If A is Al's age, then b = 9, so that 90 + c = 100 + c - f, giving f = 10,

which is impossible. Obviously A cannot be Chris's age or Bob's age. So, A is Don's

age, so that e = 9 and f = 5 + c/2. So, c must be either 0 or 2 or 4 or 6 or 8.

Now, the sum of the ages of all four = (10 + b) + (10*b + c) + 29 + (90 + 5 + c/2) = 11 *b + 3*c/2 + 134

This means that 11 *b + 3*c/2 + 134 = 200 + 10*g + c; or c/2 = 10*g- l l*(b-6)

Putting c = 0,2,4,6,8 successively, the only solution where g and b are integers <9 i sc = g = 0and b = 6. So,

Al's age = 60 years Bob's age = 29 years Chris' age = 16 years Don's age = 95 years

Chess Classic 1

There are two possible solutions.

Arrangement / Queen: R6C3 Kings: R2C3, R2C7, R6C7 Rooks: R4C1, R8C5 Knight: R8C2

Arrangement II Queen: R6C3 Kings: R2C3, R2C7, R6C7 Rooks: R4C1, R8C5 Knight: R7C1


Chess Classic 2

The two possible solutions are:

Arrangement I

Ra8, Kc5, Kd6, Rf2, Qf3, Rhl, Kh6

Arrangement II

Ra8, Kcl, Kc5, Kd6, Qf3, Rg3, Rhl

Palindromes

a) 484 = 222 is the smallest multi-digit perfect square, all of whose digits are even.

b) There cannot exist a multi-digit perfect square, all of whose digits are odd. Proof:

If there exists such a perfect square, then it must be of the form

(10*k + m)2 = 10*(10*k2 + 2*k*m) + m2 = 10*(10*k2 + 2*k*m + n) + p;

where n and p are the ten's and unit's digit of m2, and m is 1,3,5,7 or 9.

Since n is either 0 or 2 or 4 or 8 (and p is either 1 or 9 or 5), (10*k2 + 2*k*m + n) is always even, that is, the digit in the ten's place of the perfect square will always be even.

Numbers on a String

The minimum value of T = M. This minimum occurs when N = k*(BM - 1)/(B - 1), where 0 < k (an integer) < B.

Example: B = 3, M = 4

Then, N = k*(34 -1 >/(3 - 1) = 40*k, where 0 < k < 3. =e>k= 1,2

N = 40, 80


The base 3 representation of 40 is 1111. Its distinct substrings are 1, 11, 111 and 1111 (T = 4 = M). The base 3 representation of 80 is 2222. Its distinct substrings are 2, 22, 222 and 2222 (T = 4 = M).

Pirates Ahoy!

The only proportions that satisfy the conditions bulleted in the problem statement are: Two pirates: 3:4 Three pirates: 5:11:13 Four pirates: 1:19:25:28

Thus, any solution must be a multiple of one of the above proportions. The first statement 'more than 50%...' rules out two pirates and restricts the other proportions to odd multiples.

The second statement 'The least...' forces the multiple for the four-pirate proportion to be 5, 15, etc. (so that the possible solutions both have the same minimum value). If the multiple is greater than 13, then the total number of coins is greater than 1000, so the multiple can only be 5.

The last statement'... more than 10 times...' forces the solution to be that for four pirates. Given the rest of the conditions, the answer is: Four pirates with shares 5,95, 125, 140.

Polygon Trap

The new polygon formed will also be a regular polygon with 4N sides.

Now, for some properties of 2N-sided regular polygons:

If S is the length of each side and D is the distance between any two parallel sides, then S = D*tan (it/2N)

Area of a 2N-sided regular polygon = (N*S*S)/{2*tan ( J I / 2 N ) }

Also, observe that the new 4N-sided regular polygon is enclosed


in a square, whose length of each side is the distance between the parallel sides of the 4N-sided regular polygon.

Now, the length of each side of this square = S/V[4 + {tan (nl 2N)}2]

This means that the area of the new polygon = N*S*S*tan (nl 4N)/[4+ {tan (JC/2N)}2]

So, the ratio of the areas of the new polygon and the original polygon = 2*tan (n/2N)*tan (jt/4N)/[4 + {tan (Jt/2N))2]

= 2*{V (1 + x2) - 1}/(4 + x2), where x = tan (re/2N)

Zero? That's Right!

(Any permutation of the digits 2 to 9)° - 1 = 1 - 1 = 0. You may try and find a more elegant solution.

Code Cracker

If A in my code is decoded to give B, we denote it as A —> B. Then, 0123456789 ->2548719603

(0, 1,4,9,16,25,36,49,64,81,144, 169,256,625,729,784) (2,5, 7,3,59,41, 89,73,97,05, 577,593,419,941,643,607)

Russian Doll

This solution has been provided by Dheeraj Kumar Chourasia, Howrah.

Let N be the number of comparisons the person makes to fit all the dolls. By problem,

11 N 1 +2 + 3 + ... + 11 =(11 x 12)/2 = 66;or 11 N 66

Explanation: Let Nr be the number of comparisons the person makes after picking the rth doll to fit it. Then, if the rth doll is bigger than the (r -1) containers, he just compares it with the outermost and fits it (i.e. Nr = 1). However, if the rth doll is smaller than the outermost, the


person has to make another comparison and in the worst situation may have to compare till the innermost doll (i.e. N.= r -1 in the worst case) before fitting it. Hence,

1 N r r - 1; where r= 1,2,3... 12

= > ^ - 2 . 1 2 1 S r = 2 . . 2 N r ^ r = 2. .2 ( r ' 1 ) =>11 N 1+2 + 3 + . . . +12 = 66

Thus, the person makes at most 66 comparisons but certainly not less than 11 comparisons.

Got Zapped!

ZAP is 198. Here is a solution,

ZZ + AA + PP= 11 xZ + 11 xA + 11 xP = 11 (Z + A + P) => ZAP is divisible by 11.

One claim can be that Z * 0. Let us assume, on the contrary, that Z = 0. Then the equation becomes:

AP = 11 (A + P)=>A = P=>AA = 11 x2Aor, II xA = 11 x2A=> A = P = 0 This gives us the trivial case that we may ignore. So, C 1.

Next, if ZAP is divisible by 11, (P - A + Z) is also divisible by 11. As, 1 < A orZor P<9, we have-9 <-A <-1; 1<Z<9; 1<P<9.

Now, adding the inequalities we have:

- 7 S Z - A + P< 17

So, Z - A + P is divisible by 11 =>eitherZ-A + P = Oor,Z-A + P= 11.

Case /: Z - A + P = 0, then A = Z + P and ZAP = 11 (Z + A + P) = 11 (A + A) = 22A < 22 x 9 = 198.

So, ZAP is a three-digit number starting with 1 and divisible by 11. So, ZAP = 110 or 121 or 132 or 143 or 165 or 176 or 187 or 198. By inspection, ZAP = 198 is the only solution in this case.

Case II: Z - A + P = 11; or Z + P = A + 11, therefore, A + 11 = Z + P < 18 =»A<7.


Now, ZAP = 11 (Z + A + P) = 11 (2A + 11)< 11(2x7 + 11) = 275.

After considering Case I, we have,

ZAP = 209 or 220 or 231 or 242 or 253 or 264 or 275. By inspection, none of these is a solution.

Useless Cricket Statistics

1) Kumble's sets are (16, 33,48) and (48, 73,96) 2) 52

Follow Me

1 think this is the shortest solution. If not, then I have made a mistake in the design.

The idea was to design a maze in which the solution involved backtracking, returning to near the beginning of the maze. To disguise that idea, I allowed many longer solutions. In that way, the solver is encouraged to ignore the many long paths, because they all return to previously visited points. The real solution returns to previously visited points, but going in the opposite direction.

Playing With Symbols Really

The missing term is XV. The series is 1,3, 6, 10, 15, 21 written in decimal, Roman and binary notation.


Words, Words I The second, third and fourth letters of all these words form the word AND.

Words, Words II All the words AGITPROP, AGRARIAN and AGRIMONY fit the bill. All these are eight-letter words with A and G as their first two letters!

Words, Words III I have found 34 words, there may be more: IDEA • IDEAL • IDEALLY • IDEALISM • IDEALIST • IDEALISTIC • IDEALISE • IDEALISER • IDEALISED • IDEALISING • IDEM • IDENTICAL • IDENTICALLY • IDENTIC ALNESS • IDENTIFICATION • IDENTITY • IDENTIFIED • IDENTIFIER • IDENTIFYING • IDENTIFIABLE • IDENTIKIT • IDENTIFY • IDENTIFIES • IDEOGRAM • IDEOGRAPH • IDEOGRAPHIC • IDEOGRAPHICALLY • IDEOGRAPHY • IDEOLOGICAL • IDEOLOGICALLY • IDEOLOGIST • IDEOLOGUE • IDEOLOGY • IDES.

How did I get them? Simple, the last three letters of the previous word forms the first three letters, in the same order, of the following word.

Carroll Twist Here are a couple of solutions:

BEAR -> BEAT -» BOAT -> BOOT -> BLOT BLOW -> BROW -> CROW

BEAR -> BOAR -> BOOR BOOT BLOT -> BLOW BROW -> CROW


Sentence Walk

Here is a solution. Dashed lines show the path I took with arrowheads showing my movements. The resulting sentence is given below.

— r E1- » v N i r

4 - H rD-

4 /

I N i

O i +

T

• E -"R

A f J - S - - L - i H k

IE N i' • T V

+ 1 / } E t I) J -S-

i j T

•

•-F- • O- . - U

Even an Einstein found this hard

Valid Time Instant

143 such configurations are possible.

The House of Mr X

Mr X's house number is 204. To get this, we need to solve the equation n (n + 1) = 2m2, where n > m, and are both positive integers. This can be done by 'trial and error' or by more sophisticated techniques.

Gates Got It!

For the chances of showing up a larger number between pairs of dice, it can be calculated that A > B, B > C, C > D, and D > A. We find there is a non-transitive relation. Thus, the first person


cannot select a die that would give a higher chance than that of the second person.

Stack It Up

The minimum value of n is 15. First we prove that for 15 cards, the desired pair can be found. Suppose the contrary. Then cards numbered 1 and 15 must be in different stacks, as must cards 1 and 3. That means, cards 3 and 15 would be in the same stack. Therefore, cards 6 = 9 - 3 and 10 = 25 -15 are in the other stack, but which contradicts our assumption, since 6 + 10 = 16. Now we show that 14 cards can be distributed between the two stacks such that the sum of the numbers of any two cards of the same stack is not an exact square. Here is an example: 1, 2,4,6,9, 11, 13 (the first stack) and 3, 5,7, 8, 10, 12, 14 (the second stack). For any number of cards less than 14, the cards can be distributed between the two stacks in a similar way, with the desired condition holding true.

Liar Playing Spitball

Mark Martin threw the spitball.

Lunchtime Variety

Assume Kyle had the turkey sandwich. If so, then both of the first two statements would be true, which is not possible. Therefore, Kyle did not have the turkey. Now assume Neal had the turkey sandwich. If so, the both the third and the fourth statements would be true which is not possible. So if Kyle did not have the turkey and if Neal did not have the turkey, it must have been Grant. Now, assume that Kyle had the ham. This would then mean that both the second and the fourth statements were true which is not possible. So Kyle must have had the chicken.


Thus, Kyle had the chicken, Neal had the ham, and Grant had the turkey. That means, the third statement is the only true one.

Open and Shut

The state of a door depends upon its number of factors including 1 and itself. If the number of factors is even, the door is closed and if the number of factors is odd, the door is open. This is because the number of factors is actually the number of persons changing the doors state, starting from the first person opening it. For instance, let us find out the state of the 456th door, since 456 = 23*3'*19'.

Therefore, the number of factors including 1 and itself is (3 + 1 )*( 1 +1)*(1 +1) = 16. Since 16 is an even number, the 456th door is closed.

On Track

New stations: 2; Existing stations: 11.

Get the Arrangement

Here is the arrangement:

4 2 9

3 1 8

4

With ADG = 434 = 7*62, BDF = 238 = 7*34, CDE = 931 = 7*133.

On Its Head

Here is how Souvik Chaudhuri solved it. The total number of combinations possible with two-digit numbers is 20 because they are to be made up of 1, 6, 8 and 9.


Original Upside Down Difference 10 01 09 11 11 00 16 91 75 18 81 63 19 61 42 60 09 51 61 19 42 66 99 33 68 89 21 69 69 00 80 08 72 81 18 63 86 98 12 88 88 00 89 68 21 90 06 81 91 16 75 96 96 00 98 . 86 12 99 66 33

The two-digit number, which increases by the least non-zero amount when read upside down, is 86. When read so, it becomes 98, an increase of 12. (Note: 01,06,08 and 09 are not considered because they are not two-digit numbers.)

A Test for Sanity?

Alice is an insane applied mathematician. Bob is a sane pure mathematician.


Charlie is an insane pure mathematician. Dorothy is an insane pure mathematician.

Love = A White Mouse

Fluffytail has a choice: He can either start clockwise from the fifth mouse clockwise from the white mouse, or he starts anti-clockwise from the fifth mouse anti-clockwise from the white mouse.

Date of Birth

A year contains 12 months and any given month can contain at most 31 days, implying that p < 31 and q < 12 (A)

Now, the factors of 2002 (including 1 and itself) are:

1,2,7,11, 13,14,22,26,77,91,143, 154,182,286, 1001 and 2002.

But, by the problem, p and q * 1. Hence, from equation (A):

p = 2,7, 11, 13, 14, 22, 26 q = 2, 7, 11

We now analyse the situations in the light of the valid values of p and q. February 2002 (q = 2)

Date Day of the week

(P) 2 Saturday 7 Thursday 11 Monday 13 Wednesday 14 Thursday 22 Friday 26 Tuesday


July 2002 (q = 7)


(P) 2 Tuesday 7 Sunday 11 Thursday 13 Saturday 14 Sunday 22 Monday 26 Friday

November 2002 (q = 11)


(P) 2 Saturday 7 Thursday 11 Monday 13 Wednesday 14 Thursday 22 Friday 26 Tuesday

/

We recall that the date of birth of Mr de Marco in the year 2002 occurred on a Thursday. Analysing the above situations in the light of the foregoing, we observe that the following dates constitute the probable date of his birthday:

7 February; 14 February; 11 July; 7 November; 14 November

These are now analysed in the following table:


Date P q p + q Is (p + q) a perfect square

Is (p + q) even

7 February 7 2 9 = 32 YES NO 14 February 14 2 16 = 42 YES YES

11 July 11 7 18 NO YES

7 November 7 11 18 NO YES

14 November 14 11 25 = 52 YES NO

Hence, from the above it is clear that Mr de Marco's birthday occurs every year on 14 February, corresponding to p = 14 and q = 2 and he was exactly (pq)2 = 14*22 = 56 years old on 14 February 2002.

Now, 2002 - 56 = 1946. Hence, the precise date of birth of Mr de Marco is 14 February 1946.

Time Again

Defining D (x) as the remainder when x2 is divided by 26, we observe that:

S is the nineteenth letter of the alphabet and D (19) = 23 O is the fifteenth letter of the alphabet and D (15) = 17

And it is given that 'SO' corresponds to 17:23 hours. Hence, the hour relates to the second letter while the minute relates to the first letter.

Again, M is the thirteenth letter of the alphabet and D (13) = 13; Y is the twenty-fifth letter of the alphabet and D (25) = 1.

i.e. 'MY' corresponds to01:13 hours.

Now, by the problem,

A is the first letter of the alphabet and D (1) = 1

T is the twentieth letter of the alphabet and D (20) = 10

Hence, the required time corresponding to 'AT' in a twenty-four-hour watch is 10:01 hours.

X-Num and Labyrinths

X-Num

If you enjoy crosswords, here is something for you. In this case, however, instead of word clues you get number clues.

1 2 • 5 6 7

8 9

10 11

12 13

14 15 16 17 • • • 20 21

22 • * • • • 1 26 • 1 • 3 m •

Figure 18

X - N u m a n d L a b y r i n t h s 199

Across

I Add one to the smallest number that is the sum of three different cubes in two ways and you get this answer 4 Starting from this three-digit number to 831 is the largest gap between two semi-primes less than 1000 6 The fifth prime 5 A six-digit number of the form abcdef where abc is the smallest number that appears in its factorial six times; ef is the last two-digit string to appear in the decimal expansion of ; and f is the number of Platonic solids 9 When increased by four, this three-digit number becomes a cube 10 Same as 1 across II The last digit of this three-digit number is 4; the first two form a two-digit number that is the fifth Catalan number 13 A four-digit number of the form abed, not all digits are different; ab is the smallest number that is not of the form [2X -3y |; cd is the smallest number n with the property that there are no numbers relatively prime to n smaller numbers 14 A six-digit number of the form abcdef, not all digits are different, and abc is the sum of the first three fifth powers; def is a multi-perfect number 18 A number of the form aba, where a = 5 and ba is the number of hexominoes 19 This five-digit number when reversed remains unchanged, and the same is true about its mirror image 22 A sum of the first five fourth powers 23 A three-digit number of the form aba where a is the largest cube in the Fibonacci sequence; and b is the only even prime 24 The minimum sum of pandigital three-digit primes 25 It isa 'strbogrammatic year' between 1691 and 1961,and the next such year will be 6009 26 This number is a palindrome in both binary and base 10


27 An eight-digit palindrome for abbbbbba; where a is the first exact number as defined by Euclid and b is the only cube that is one less than a square

Down

1 The first and the second digits of this five-digit number make a two-digit number that is the only example of a square that is two less than a cube; the third digit is 2, and the fourth and the fifth digits are first and second digits written in the reverse order 2 1111 in base 8 3 The first two digits of this three-digit number make a two-digit number ab and (ab)2 is a palindrome; the third digit of the number is 2 4 Itself a repeated digit, and its square ends in a repeated digit 5 Twin prime of 149 6 A number of the form abed, not all different digits, such that ab is the smallest abundant number and cd is the smallest number of distinct squares needed to tile a square 7 A number of the form aabbaa, such that aa is the largest known multiplicative persistence; bb denotes the number of derangements of 5 items. A derangement of 'n' ordered objects, denoted' !n\ is a permutation in which none of the objects appear in their 'natural' (i.e. ordered) place. For example, the only derangements of (1, 2, 3) are (2, 3, 1) and (3, 1, 2), so, !3 = 2 9 Put a 3 at the end and you get a six-digit prime greater than 141403 12 A three-digit number such that the first and second digit forms the sixth tetrahedral number and the second and the third digits form magic constant in a 5 by 5 magic square 15 The first digit of this three-digit number is 7 and the second and the third digits give us approximately the number of degrees in a radian


16 A six-digit number abedef such that a + b = c = d = e + f; a = f, b = e and none of the digits is 0. Also, ef forms a two-digit number that is the first number after 1 to be both square and triangular 17 Boiling point of water at degree Fahrenheit 20 A number of the form aabaa where aa is the only number known whose square has no isolated digits; and b is the multiplicative identity 21 A six-digit number of the form ababa, where a3 + 1 = b and a = b - 1; also a and b are not 0 22 A four-digit number of the form abba such that a = b2; bb is two-digit, is semi-prime and is the largest number that is not the sum of distinct triangular numbers 23 A four-digit number of the form abba such that ab = a1 + b2

Make Your Own Maze

lake a squared paper and draw out a rectangle with an odd number of squares on each side. This design has 19 squares on cach side, but it could be a rectangle. When drawing the maze, use a soft pencil, as you'll be rubbing lots out!

Figure 24


Fill in alternate lines and columns, like the picture above, making a waffle pattern. All the white squares will end up as part of the paths. All the black squares will be walls. Grey squares will be either paths or wall. Shade the grey squares very lightly, as you will be rubbing some of them out.

• •

• • • • -•

• • • • •

Figure 20

Now, you need to choose an entry point to the maze at its edge (choose a grey square) and rub it out. Similarly, choose a destination (a grey square, again). For my maze, I've chosen an exit on the edge, but you can have a destination in the middle of the maze. Rub out the chosen square. The remaining squares on the outermost boundary should be coloured black to represent the outer wall of the maze.

Next, draw the main path, or what will actually be the solution to the maze. Start from the entrance and, rubbing out only grey squares, make a path to the final destination. Make sure that the path is not too short (too easy to solve!) or too long, for then you won't be able to have many choices for dead ends and it wouldn't be a very challenging maze. It's quite fun to head for the exit and then turn away again.

Wiggle the path around, but make sure that the path never crosses itself. This can be surprisingly tricky. If your path does


meet itself, redraw a few grey walls and try again. (I had to do this a few times while drawing this maze.)

• • • • • • • • • • • • • • • • • • • • • I I I • • • • • • • •

• • • • • • • • • • • • • • • •

Figure 21

You will have some white squares left that aren't connected to the main path. So you convert them to dead ends off the main path. Find a white square all by itself. Make a path (remember, only rub out grey squares) towards the main path. This is now a dead end.

You can have one dead end branching off another, like the small spiral in the bottom left-hand comer. Try to make some dead ends look more attractive than the main path, such that the dead end heads towards the destination, while the main path snakes away from it! You now have the finished design, so all you need to do is to tidy it up a little.

• • • • • • • •

• • • • • • • • • I | | • • • • •

• • mm _

• • • • • • • I | | • • • • •

• • • • • • • I | | • • • • •

— H

• • • • • • • • • • •i • • • • • •

• • • • •

H m • • • • •

• • • • • • •

Figure 24


Change all the remaining medium pale squares into black, as they are now officially walls. Remove all the lines outlining the squares (or if this is too boring, leave them in, and pretend that it's a Roman mosaic!). I've made the paths look the same as the background.

Figure 23

This type of maze is possible to solve even if you don't know the design. Imagine that you are entering the maze, and put your right hand on the wall on the right. Walk forwards, but whenever there is a path to your right, turn right. This will keep your hand moving along the side of the wall. If you meet a dead end, turn round, and your right hand will now be on the other wall. Eventually, you will reach the final destination (although you may go down several dead ends). In fact, if you turn round at the destination, and re-enter the maze, still keeping your right hand on the wall, you will travel through the rest of the maze, and get back to the entrance, having travelled through every passageway twice (but not necessarily in the same order), and touched every piece of wall. You can, of course, use your left hand instead of your right.

If you take this sort of maze, and colour red all these walls which you touch with your left hand as you go from entrance to


exit, and colour green all the walls that you touch with your right hand, you will find out that the direct path has walls of both red and green. All dead ends have only red or only green walls.

Many modern mazes are island mazes, which are more complicated than this, with paths that do cross each other. They can't be solved in the same way, as you just go round and round! Also, they don't have a single solution. You can easily make such a maze, by designing a maze as above, and then removing a few more walls. This has apparently been done with the I lampton Court Maze. Don't remove too many, else you'll make the maze too easy! (Jo Edkin)

Cluzzle

You have to get out of the following maze. Enter it from square A and exit from square \|/. You have been given thirty-six hints and they are all in the form of puzzles or teasers. And, the answers to all the clues are numbers. Answers to some of these clues could be the same, and the path out of this maze is to be found using these squares.

A B c D E F

G H I J K L

M N o p Q R

S T u V W X

Y Z n (E r A

T c 7> 0 A V

Figure 24


A: The number of days in a Lunar cycle B: The smallest number with 7 divisors C: How many teeth does a normal human adult have? D: In degrees Fahrenheit, at what temperature does water freeze? E: The largest number that cannot be written as a sum of numbers whose reciprocals sum to 1 F: The smallest number that can be written as the difference of two squares in four ways G: Starting with 12,496, how many links does the longest known sociable chain have? H: The number of dominoes in a standard double-six set I: This is the smallest number of colours sufficient to colour all planar maps J: The fifth power of 2 K: The number of commutative semi-groups of order 4 L: Which is the largest number divisible by all numbers less than its square root? M: What is the smallest number of distinct squares needed to tile a square? N: In the Imperial system of weights and measures, how many pounds make a quarter? O: The first triangular number that is a sum of two cubes P: The number of essentially different placements of eight non-attacking queens on a chessboard Q: Which is the smallest number with the property that its first five multiples contain the digit 9? R: It has the property that placing the last digit first gives 1 more than its triple


S: The smallest number that is the product of two different substrings I: The smallest number n, where either n or n + 1 is divisible by numbers 1 to 8 11: According to Descartes, this is an odd perfect number less than 100. He defined such numbers as a product of a square and a prime V: The average number of days the menstrual cycle lasts W: How many countries were there in the Allied coalition in the Gulf War of 1991? X: Which is the smallest number to appear six times in Pascal's triangle? Y: The smallest number with the property that its first three multiples contain the digit 2 /.: The largest number, which cannot be written as the sum of distinct primes of the form 6n + 1 11: The smallest number with the property that its first three multiples contain the digit 7 (I•: The smallest number with twenty divisors I : Acturus is the fourth brightest star in the sky and is times larger than the sun A: The smallest number which is palindromic in bases 5 and 6 I The sum of the first six triangular numbers C,: How many horizontal and vertical lines do you find on a Go board? b: A pyramidal number and triangular number less than 60 (-): Atomic number of iron A: The number digits in the Pharaoh's cubit v|/: The seventh triangular number


Solutions

X-Num

Cluzzle

The common number is 28 and the path is as shown below:

A T B c D E F

G ^ H ? i J K L

M N ^ O T p Q R

S T W T X

Y Z n (E A

T C T> 0 A ^

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