greking: vedic maths concept

Target 320+ in GRE

1

GREKing

Vedic Numbers

Number Tree LCM HCF Divisibility Rules Power cycle Remainder Theorem Remainder of powers an – bn

Last and Second last digit Power of Exponents Euler’s Theorem Fermet’s Theory Wilson Theorem Number Systems (decimal binary)

Importance in exams

CAT – 4 questions

CMAT – 2 questions

NMAT – 5 questions

Number Tree

HCF & LCM

42 = 21 x 2= 7 x 3 x 2

• Factorize the following numbers: • 42 • 72 • 84 • 65 • 108• 210

• Calculate the LCM and HCF for the following groups of numbers

a) 42, 70

b) 18, 24, 60 ,150

LCM of nos 60

• Three bells ring after 3, 4 and 5 minutes respectively. If they start ringing together, when

• will they ring together again?.

a) 50b) 90c) 60d) 42

HCF of nos 10

• Three buckets of milk having capacities 40, 30 and 50 L respectively. What is the largest

• size of the measure that can be used to pour milk from all three buckets such that the

• volumes of milk contained in each bucket is an integral number of pourings used by the

• measure.

a) 50b) 10c) 60d) 42

A = 12 xB = 12 y144 xy = 2880xy = 20

1, 202, 104, 55,410,220,1

• The product of two numbers is 2880. If the HCF is 12, how many such pairs of number are possible?

a) 3b) 6c) 8d) 2

LCM of nos720

• Find the least number which when divided by 9,12,16,30 leaves in each case a

• remainder of 3.

a) 721b) 722c) 723d) 754

LCM of nos 15105120135…..

• Find the number of all the numbers divisible by 3 and 5 between 100 to 200.

a) 9b) 8c) 6d) 7

M(3) + M(5) – M(3,5)30 + 20 – 7= 43

• Find the number of all the numbers divisible by 3 or 5 between 100 to 200.

a) 25b) 89c) 60d) 43

LCM of nos 15105120135…..

Sn = n/2 (a+l)

• Sum of all the numbers divisible by 3 and 5 between 100 to 200.

a) 1050b) 8090c) 6020d) 4231

120• What is the least three digit number that is

divisible by 5 and 6?

• - What is the least 4 digit number that is

divisible by 12 and 15?

• - What is the least 4 digit number that is

divisible by 25 and 30?

Rules of Divisibility

TESTS OF DIVISIBILITY:

• Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.

• Ex. 84932 is divisible by 2, while 65935 is not.

• Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.

• Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3.

• But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.

Last 3 multiple of 8

• Find the value of K when 425K is divisible by 8..

a) 6b) 5c) 0d) 4

Rule of 2 than 9

• A number A4571203B is divisible by 18. Which of the

• following values can A and B take ?

a) 1, 2

b) 2, 3

c) 6, 8

d) 3, 3

Rule of 8 and 11

• Find A and B , if 3765A56682B is divisible by 88?

21 = 222 = 423 = 824 = 1625 = 32……

24

86 4k

Power Cycle

4k = 2, 3, 7, 82k = 4, 91k = 1, 5, 6, 0

• 1 = 1, 1, 1, …• 2 = 2, 4, 8, 6…• 3 = 3, 9, 7, 1…• 4 = 4, 6, 4, 6• 5 = 5, 5, 5, 5…• 6 = 6, 6, 6, 6…• 7 = 7, 9, 3, 1…• 8 = 8, 4, 2, 6…• 9 = 9, 1, 9, 1…• 0 = 0, 0, 0, 0

• What is the last digit of 360

• What is the last digit of 743

• What is the last digit of 940

• What is the last digit of 56120

• What is the last digit of 26670

• Find the unit's digit in

• (264)102 + (264)103

a) 0b) 2c) 4d) 20

NMAT• Find the unit's digit in

• (263)102 x (265)103

a) 5b) 2c) 4d) 20

Remainder Theorem

• Rem (X / A) = Rem (Y/A) x Rem (Z/A)

• Where y and z are factors of X

• Find the remainder of

a) 24^3 / 7

b) 72^6 / 12

c) 625^12 / 12

• Find the remainder when 2^246 is divided by 7.

a) 1b) 3c) 7d) 9e) none of these

3^2720

• Find the right most non zero digit of (30)2720

• a.1 b.3 c.7 d.9.

Euler’s Theorem

N + 1 / N is always 1N – 1 / N is always 1, -1

• Find the remainder when 3560 is divided by 2.

a) 5b) 2c) 4d) 1

N + 1 / N is always 1N – 1 / N is always 1, -1

• Find the remainder when 9560 is divided by 2 =

• 33560 is divided by 2 =

• 5560 is divided by 4 =

N – 1 / N is always 1, -1

• Find the remainder when 7560 is divided by 2 = • 31560 is divided by 2 =

• 3560 is divided by 4 =

16 / 17 = -1So 2 – 1 = 1

CAT 2002• Find the remainder when 2256 is divided by 17 =

a) 3b) 1c) 7 d) 9.

2.2256

16 / 17 = -1So 2 – 1 = 1

• Find the remainder when 2257 is divided by 17 =

A. 3B. 2 C. 7D. 9

Cyclic Remainders

541 / 7 = 5542 / 7 = 25 / 7 = 4543 / 7 = 20 / 7 = 6544 / 7 = 2545 / 7 = 3546 / 7 = 1

XAT LEVEL• Find the remainder when 54120 is divided by 7 =

a) 3b) 1c) 7d) 9.

41 / 6 = 442 / 6 = 4……

CAT 2003• Find the remainder when 496 is divided by 6 =

a) 3b) 4c) 7d) 9

•An+Bn

• Is always divisible by A + B for all odd n

•An - Bn

• Is always divisible by A - B for all odd n• Is always divisible by A - B for all even n• Is always divisible by A + B for all even n

351 - 231 = 12

SNAP• Find the remainder when• 3523 - 2323 is divided by 12

a) 3

b) 0

c) 7

d) 9

352 - 232 = 12, 58

• Find the remainder when• 3522 - 2322 is divided by 58

a) 3b) 0c) 7d) 9.

7+6 = 13;73 – 63 = 127

CAT 2002• Find the remainder when• 76n – 66n is divided by ?

a) 13b) 127 c) 559d) All of these

Highest powers in n!

780/7 + 780/49 + 780/343

CMAT• Find the highest power of 7 which will divide 780!

a) 13

b) 127

c) 559

d) 15

250/5 + 250/25 + 250/125

• Find the highest power of

• 25 which will divide 250!

a) 13

b) 127

c) 559

d) 31

Wilson's Theorem

• Let p be an integer greater than one. p is prime if and only if (p-1)! = -1 (mod p).This beautiful result is of mostly theoretical value because it is relatively difficult to calculate (p-1)! In contrast it is easy to calculate ap-1, so elementary primality tests are built using Fermat's Little Theorem rather than Wilson's. Neither Waring or Wilson could prove the above theorem, but now it can be found in any elementary number theory text. To save you some time we present a proof here.

• Proof. It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is composite, then its positive divisors are among the integers1, 2, 3, 4, ... , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p).

• However if p is prime, then each of the above integers are relatively prime to p. So for each of these integers a there is another such that ab = 1 (mod p). It is important to note that this b is unique modulo p, and that since p is prime, a = b if and only if a is 1 or p-1. Now if we omit 1 and p-1, then the others can be grouped into pairs whose product is one showing2.3.4.....(p-2) = 1 (mod p) (or more simply (p-2)! = 1 (mod p)). Finally, multiply this equality by p-1 to complete the proof.

Vedic Numbers

Rem. Of(17/12*19/12*21/12)

• Let N=1421*1423*1425.What is the remainder when N is divided by 12?

a) 0b) 9c) 3d) 6

2^4+2/5=3.

• What is the remainder when (2^100 + 2) is divided by 101?

a) 3b) 4c) 5d) 6

2^n & 5^n=n+1.

• The values of numbers 2^2004 and 5^2004 are written one after another. How many digits are there in all?

a) 4008b) 2003c) 2005d) none of these

3^2720=1.

• Find the right most non zero digit of (30)^2720.

a) 1

b) 3

c) 7

d) 9.

A^n + B^n=divisible by a+b

• The remainder, when (15^23 + 23^23) is divided by 19 is:

a) 4

b) 0

c) 15

d) 18.

Divisibility rule of 27

• What is the remainder of 22222222222….(27 times) is divided by 27?

a) 0

b) 1

c) 3

d) 4

N^n+1 /n=1

• Find the remainder when 2009^2010 is divided by 2011.

a) 2010

b) 0

c) 1

d) 2009

Understand the format of Q

• 111^111 +222^111 +333^111 +444^111+….999^111 is divided by 555?

a) 1b) 0c) 2d) 3e) 4

1! +2*2!/3=rem. is 2

• Let N!=1*2*3*4*5*6*N for N greater or equal to 1.

• If P=1! +(2*2!) +(3*3!) +(4*4!)……(12*12!). Find the

remainder when P is divided by 13.

a) 11

b) 1

c) 2

d) 12

• If R=30^65 -29^65/30^64+29^64. then,

a) 0<R<1

b) 0.5<R<1

c) 0<0.1

d) R>1

306 is divisible by 3, 6, 9

• 3066 - 306 is not divisible by which of the following ?

a) 3

b) 4

c) 6

d) 9

We need 1 even and two odd

• P,Q& R form a set if distinct prime numbers less than 20.How many such prime numbers are possible for which the sum of P,Q & R is even.

a) 15b) 21c) 24d) 28

• Q5. (153)8 = (x)10.

a) 162b) 107c) 166d) 207e) 203

Find the last two digits if 1733^148.

• Finding the last n second last digit

• (2^10)even= 24

• (2^10)odd =76

• Finding the last n second last digit

• For odd digits

• (21)^27= 41

• (31)^21=31

• (a1)^cd = ad1

Concept of perfect squares examining the nature of all perfect squares…

Any number will have a unit’s digit as..1,2,3,4,5,6,7,8,9,0. Examining the nature of 1.

1^2=0111^2=12121^2=44131^2=961 n so on…

Nature of 2.2^2=0412^2=14422^2=484 32^2=1024 n so on…

Nature of 3.

3^2=0913^2=16923^2=52933^2=1089..

Nature of 4

4^2=1614^2=19624^2=57634^2=1156..

Nature of 5

5^2=2515^2=22525^2=62535^2=1225..

Nature of 66^2=3616^2=25626^2=576..

Nature of 7

7^2=4917^2=28927^2=729..

Nature of 8

8^2=6418^2=32428^2=784..

Nature of 9

9^2=8119^2=36129^2=841..

74= 2401 => last digit 1 78= 5764801 => last digit 1 72008 = 1 , cyclist of four.

What is the last two digits of 72008

a) 71

b) 51

c) 01

d) 11

X can be even or odd & same for Y.

If x2 - y2 = 1234Find the number of integral solutions of x,y.Where x, y < 150.

a) 4

b) 5

c) 0

d) 8

To get the last digit as 9,the only ways are

1&8,2&7,3&6,4&5,9&0

If x2 + y2 = 3479Find the number of integral solutions of x,y.Where x, y < 300.

a) 4

b) 5

c) 0

d) 8

ASHITA MEPANIBOSTON UNIVERSITY

MOHIT DESAISYRACRUSE UNIVERSITY

FATEMA AURANGABADINORTHEASTERN UNIVERSITY

BHAVISHA DAWDANEW YORK UNIVERSITY

MAITRI BUCHUNIVERSITY OF TEXAS DALLAS

RAVI HARIANITUSCON, ARIZONA

SANKET WALAWALKARKELLY SCHOOL OF BUSINESS

AMEY BHAVSARNEW YORK UNIVERSITY

KUSHAL THAKKARCARNEGIE MELLON UNIVERSITY

ANUP DESHPANDENEW JERSEY, NOMURA

ROHIT ZAWARINDIANNA UNIVERSITY

SAGAR SUCHAKUNIVERSITY OF TEXAS DALLAS

DARSHI SHAHNEW YORK UNIVERSITY

AAMIR LAKDAWALAKAISERSLAUTERN UNIVERSITY, GERMANY

320 + scoring students from GRE King

ARCHANA RAMAKRISHNANUNIVERSITY OF CALIFORNIA

Target 320+ in GRE

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