RMO or Pre-RMO training: Homibhabha Science Foundation exam

Sample questions based on logic only:

  1. Given the sixty-four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position). Show that alll the sixty four entries are in fact equal.
  2. Let T be the set of all triples (a,b,c) of integers such that 1 \leq a <b<c \leq b. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in F. Prove that the sum is divisible by 7.
  3. In a class of 25 students, there are 17 cyclists, 13 swimmers and 8 weight lifters and no one is all the three. In a certain mathematics examination, 6 students get grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C determine the number of students, who got grade A. Also, find the number of cyclists who are swimmers.

Any ideas ? Please wait for solutions tomorrow.

Nalin Pithwa.

A tough nut from Norway: RMO training !


The integers 1, 2, 3, 4, and 5 are written on a blackboard. It is allowed to wipe out two integers a and b and replace them with a+b and ab. Is it possible, by repeating this procedure, to reach a situation, where three of the five integers on the blackboard are 2009?


The answer is NO.

First notice that in each move, two integers will be replaced with two greater integers (except in the case where the number 1 is wiped out). Also note that from the start there are three odd integers. If one chooses to replace two odd integers on the blackboard, the number of odd integers on the blackboard decreases. If one chooses to replace two integers, which are not both odd, the number of odd integers on the blackboard is unchanged. To end up in a situation, where three of the integers on the blackboard are 2009, then it is not allowed in any move to replace two odd integers. Hence, the number 2009 can only be obtained as a sum of a+b.

In the first move that gives 2009 on the blackboard, two integers a and b are chosen such that a+b=2009 and either ab>2009 or ab=2008. In the case, ab=2008, one of the two chosen numbers is one; and hence, 1 will no longer appear on the blackboard. In either case, the two integers a+b=2009 and ab that appear in the creation of the first 2009 cannot be used anymore to create new instances of 2009. The second 2009 can only be obtained by choosing c and d such that c+d=2009 and either cd>2009 or cd=2008, and just as before; the numbers c+d=2009 and cd cannot be used in obtaining the last 2009. So, after forming two instances of 2009, there are four integers on the blackboard that have become useless for obtaining the third instance. Hence, the last integer 2009 cannot be obtained.


Ref: Nordic Mathematical Contest, 1987-2009.

Nalin Pithwa.

RMO Training: more help from Nordic mathematical contest


32 competitors participate in a tournament. No two of them are equal and in a one against one match the better always wins. (No tie please). Show that the gold, silver and bronze medal can be found in 39 matches.


We begin by determining the gold medallist using classical elimination, where we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners in the second round, then 2-semifinal matches and finally one match making 31 matches altogether.

Now, the second best player must have at some point lost to the best player, and as there were 5 rounds in the elimination, there are 5 candidates for the silver medal. Let C_{i} be the candidate who  lost to the gold medalist in round i. Now, let C_{1} and C_{2} play, the winner play against C_{3}, and so forth. After 4 matches, we know the silver medalist; assume this was C_{k}.

Now, the third best player must have lost against the gold medalist or against C_{k} or both. (If the player had lost to someone else, there would be at least three better players.) Now, C_{k} won k-1 times in the elimination rounds, the 5-k players C_{k+1}\ldots C_{5} and if k is greater than one, one player C_{j} with j<k. So there are either (k-1)+(5-k)=4 or (k-1)+(5-k)+1=5 candidates for the third place. At most 4 matches are again needed to determine the bronze winners.

Cheers to Norway mathematicians!

Nalin Pithwa.

Reference: Nordic mathematical contests, 1987-2009.

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RMO Training: taking help from Nordic mathematical contest: 1988


Let m_{n} be a smallest value of the function f_{n}(x)=\sum_{k=0}^{2n}x^{k}. Prove that m_{n} \rightarrow \frac{1}{2} when n \rightarrow \infty.


For n>1,


From this, we see that f_{n}(x)\geq 1 for x \leq -1 and x\geq 0. Consequently, f_{n} attains its maximum value in the interval (-1,0). On this interval


So, m_{n} \geq \frac{1}{2}. But,

m_{n} \leq f_{n}(-1+\frac{1}{\sqrt{n}})=\frac{1}{2-\frac{1}{\sqrt{n}}}+\frac{(1-\frac{1}{\sqrt{n}})^{2n+1}}{2-\frac{1}{\sqrt{n}}}

As n \rightarrow \infty, the first term on the right hand side tends to the limit \frac{1}{2}. In the second term, the factor


of the numerator tends to zero because

\lim_{k \rightarrow \infty}(1-\frac{1}{k})^{k}=e^{-1}<1.

So, \lim_{n \rightarrow \infty}m_{n}=\frac{1}{2}

auf wiedersehen,

Nalin Pithwa.

Reference: Nordic Mathematical Contest, 1987-2009.


RMO Training: a question from Nordic Mathematical Contest: 1988


Two spheres with the same centre have radii r and R, where r<R. From the surface of the bigger sphere we will try to select three points A, B and C such that all sides of the triangle ABC coincide with the surface of the smaller sphere. Prove that this selection is possible if and only if R<2r.


Assume A, B and C lie on the surface \Gamma of a sphere of a radius R and centre O, and AB, BC, and CA touch the surface \gamma of a sphere of radius r and centre O. The circumscribed and inscribed circles of ABC then are the intersections of the plane ABC with \Gamma and \gamma, respectively. The centres of these circles both are the foot D of the perpendicular dropped from O to the plane ABC. This point lies both on the angle bisectors of the triangle ABC and on the perpendicular bisectors of its sides. So, these lines are the same, which means that the triangle ABC is equilateral, and the centre of the circles is the common point of intersection of the medians of ABC. This again implies that the radii of the two circles are 2r_{1} and r_{1} for some real number r_{1}. Let OD=d. Then, 2r_{1}=\sqrt{R^{2}-d^{2}} and r_{1}=\sqrt{r^{2}-d^{2}}. Squaring, we get R^{2}-d^{2}=4r^{2}-4d^{2}, 4r^{2}-R^{2}=3d^{2} \geq 0, and 2r\geq R.

On the other hand, assume 2r \geq R. Consider a plane at the distance d=\sqrt{\frac{4r^{2}-R^{2}}{3}} from the common centre of the two spheres. The plane cuts the surfaces of the spheres along concentric circles of radii

r_{1}=\sqrt{\frac{R^{2}-r^{2}}{3}} and R_{1}=2\sqrt{\frac{R^{2}-r^{2}}{3}}.

The points A, B, and C can now be chosen on the latter circle in such a way that ABC is equilateral.

Reference: Nordic Mathematical Contest, 1987-2009.


Nalin Pithwa.

An easy inequality from Nordic mathematical contests !?

Reference: Nordic Mathematical Contest, 1987-2009, R. Todev.


Let a, b, and c be real numbers different from 0  and a \geq b \geq c. Prove that inequality

\frac{a^{3}-c^{3}}{3} \geq abc(\frac{a-b}{c} + \frac{b-c}{a})

holds. When does the equality hold?


We know that a, b and c are real, distinct and also non-zero and also that a \geq b \geq c.

Hence, c-b \leq 0 \leq a-b, we have (a-b)^{3}\geq (c-b)^{3}, or

a^{3}-3a^{a}b+3ab^{2}-b^{3} \geq c^{3}-3bc^{2}+3b^{2}c-b^{3}

On simplifying this, we immediately have

\frac{1}{3}{(a^{3}-c^{3})} \geq a^{2}b-ab^{2}+b^{2}c-bc^{2}=abc(\frac{a-b}{c}+\frac{b-c}{a}).

A sufficient condition for equality is a=c. If a>c, then (a-b)^{3}>(c-b)^{3}. which makes the proved inequality a strict one. So, a=c is a necessary condition for equality too.

-Nalin Pithwa.