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Pre RMO Hints and Solutions

1. Score of 89, 88 and 85 are impossible to score as to score 88 or 89, one needs to get morethan 29 questions correct and atleast one question wrong, whereas to score 85 one needs to

get more than 28 questions correct with two questions wrong which is impossible as there

are only 30 questions.

2. = (( ) ( ) ( )) . Thus for it to be equalto zero, all the numbers must be equal which is not possible. If two numbers are equal, then

the expression can be equal to 1 ( = = 1, = 2) or is greater than3 (Take = = 1, = 3). If all three are distinct, then their differences are atleast 1, 1 and2 (Take = 1, = 2, = 3) giving the least value of the expression in this case as 3. Thusthe expression cannot be equal to 2 either.

3. As 3 4 = 2013, we get = ()

, therefore must be a multiple of3. As ispositive, we get = () > 0, or 503.25 > . Thus takes the values 3,6,9, ,501,i.e. total 167 values. As for every such value of, there is one positive integer value of, weget 167 such pairs.

4. Let = 10 , thus we get two equations for two variable and as 10 =2(10 ) 7 and 10 = 7( ) 6. Solving simultaneously, we get = 8 and = 3.

5. If we let = , then = , = , = , , = , = , i.e. = for odd valueof and = for even value of. Therefore if = 4, we must have two of , , , as and the other two as since is rational. Therefore two of,, , areodd and the other two are even. Therefore the product must be divisible by 4. Thus 945 270 210. Also for { ,, ,} = {2,4,5,7}, we get = 280.

6. Observe that 3 = 243 = 61 4 1, thus 3 = (3) = 61 1 for some integer .Therefore 3 = (3) = (61 1) (61 1) (61 1) = 61 1 forsome integer . i.e. the remainder if3 is divided by 61 is 1. Now 3 = 243 3 =(6 1 3 6 0) (61 1) = 61 60 for some integer . i.e. the remainder when

3

is divided by 61 is 60. Similarly deduce that the remainder when 11

is divided by61 is also 60 by noting 11 = 61 2 1 and writing 11 as 11 11 1 1(1007)times to get 11 = 61 1 = 61 ( 1) 60 for some integer . Thus the requiredremainder is obtained by dividing 60 60 = 120 by 61 giving = 59.

7. Take = 123 and = 133 to write 256 as . Therefore 256 123 133 =( ) = 3( ) = 3 123 133 256.

8. Let = ( 1) () . Get the values of and by putting = 1 and 1.9.

1is the LCM of first ten natural numbers. (Here the number 1 is excluded from dividing

N or else the question wouldnt have made sense).

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10.The perfect squares are 2, 4, 6,,100 as well as 1, 9, 25, 49, 81.11.Let the altitudes meet in (They are concurrent). Then the quadrilaterals and

are cyclic (Why?). Therefore = = 90 . Similarly = = 90 (Why?).

12.Get the remainder as ( 1) ( 7) by carrying out long division. For the remainder tobe zero we must have = 1, = 7.

13.Drop perpendicular on side meeting at the point . Thus = = 6 and =8. Now ~. Thus =

i.e.

=

/ or 2 3 = 12.

14.Draw parallel to side from the point meeting the side in . Thus = = 4(Why?) and as well as =

= 3 (Why?). Since is right angled, = 5.

15.Observe that = 1 and = 1, thus + = and + = giving + = . As = = 2 and = = 1, we get = , = , = and so on.

16. = 111111 = 3 7 37 11 13 = ( ) ( ) ( ). Compare to get = 3, = 7 and = 4.

17.Assume = 9 45 9 45 , thus = 18 3 or ( 3)( 3 6) = 0 or = 3.

18.For = 41, 82, 123, the expression is divisible by 41. (But it takes prime values for =1,2,3,,39). !

19.Observe that 1 = ( 2 1 ) = ( 1) = ( 1 )( 1 ), then proceed.

20.Join and and use midpoint theorem.21. = 1 is the identical root (Obtained by hit and trial method).22.The quadrilateral is cyclic (Why?). Thus = = = 45 (Why?). As

= 90, = 45, Thus = = 6.

23. is the midpoint of. Let = . Drop perpendicular from to to complete a righttriangle. Use Pythagoruss Theorem to get ( 5) = 15 (3 0 ), i.e. = 22.

24.4( ) 12( ) 27 = 0 or (2 3) (2 3) (2 3) = 0giving = = = .

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25.

= 2 = 121, thus = 11.

= 2 = 13

9. Thus = 13 13.

26.Let = 3, thus = 4 (Why?). Use Pythagoruss theorem to get = 7.27.Add all the three equations to get 6 2 4 = 14 or ( 3)

( 1) ( 2) = 0, i.e. = 3, = 1, = 2.

28.Assume the radii or the three circles to be , , . Solve the simultaneous equations to getthe radii as 1,2,3.

29.Assume 2 = 3 = 6 = , therefore = 2, = 3 and = . Thus +

+

= 2 3 =

1 to get

= 0. (Here it was assumed that ,, are non-zero or otherwise the

expression wouldnt have been defined.)

30.Put = to get ( 1) = ( 1) or 3 2 3 = 0 or 3 2 3 = 0.Thus

=

=

+ =

.

Authors Comments on the paper

Itll be almost impossible for any student to solve all or nearly all questions in the time given,

however that is not at all expected. What is expected is the students ability to think beyond

textbooks and the will to lookout for challenging problems which precisely is the goal of Olympiads.

A student is not expected to know all the techniques used here to solve a particular question, but

with a little innovative thinking and with some patience and time, many questions should be doable

within the knowledge acquired at the school level.