# Pre RMO or RMO Solutions: Homi Bhabha Science Foundation

Problem:

Reference: Problem Primer for the Olympiad by C. R. Pranesachar, et al. Prism Books

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his own cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Solution:

Suppose E is wearing a white cap:

Then, D is lying and hence must be wearing a white cap. Since D and E both have white caps, A is lying and hence, he must be wearing white cap. If C is speaking truth, then C must be wearing a black cap and B must be wearing a black cap as observed by C. But, then B must observe a black cap on C. Hence, B must be lying. This implies that B is wearing a white cap which is a contradiction to C’s statement.

On the other hand, if C is lying, then C must be wearing a white cap. Thus, A, C, D and E are wearing white caps which makes B’s statement true. But, then B must be wearing a black cap and this makes C statement correct.

Thus, E must be wearing a black cap. This implies that B is lying and hence, must be having a white cap. But, then D is lying and hence, must be having a white cap since B and D have white caps. A is not saying the truth. Hence, A must be wearing a white cap. These together imply that C is truthful. Hence, C must be wearing a black cap. Thus, we have the following distribution:

A: white cap; B: white cap; C: black cap; D: white cap; E: black cap.

Hope you enjoyed it! There can be some other approaches too starting with some other assumption(s).

Nalin Pithwa.

# Pre-RMO or RMO sample problem solutions: logic questions

Reference: Problem Primer for the Olympiad by C. R. Pranesachar et al, Prism Books.

Problem 1:

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 all the sixty four entries are in fact equal.

Solution 1:

Consider the smallest value among the 64 entries on the board. Since it is the average of the surrounding numbers, all those numbers must be equal to this number as it is the smallest. This gives some more with the smallest value. Continue in this way till all the squares are covered.

Problem 2:

Let T be the set of all triples $(a,b,c)$ of integers such that $1 \leq a < b . For each triple $(a,b,c)$ in F, take the product abc. Add all these products corresponding to all triples in T. Prove that the sum is divisible by 7.

Solution 2:

For every triplet $(a,b,c)$ the triplet $(7-c,7-b,7-a)$ is in T and these two are distinct as $7 \neq 2b$. Pairing off $(a,b,c)$ with $(7-c,7-b,7-a)$ for each $(a,b,c) \in T$. 7 divides $abc-(7-c)(7-b)(7-a)$.

Problem 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 got 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.

Solution 3:

Let S denote the set of all 25 students in the class, X the set of swimmers in S, Y the set of all weight lifters and Z the set of all cyclists. Since students in $X \bigcup Y \bigcup Z$ all get grades B and C and six students get grades D or E, the number of students in $X \bigcup Y \bigcup Z \leq 25-6=19$. Now assign one point to each of the 17 cyclists, 13 swimmers, and 8 weight lifters. Thus, a total of 38 points would be assigned among the students in $X \bigcup Y \bigcup Z$. Note that no student can have more than two points as no is all the three. Then, we should have $|X\bigcup Y \bigcup Z| \geq 19$ as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in $X \bigcup Y \bigcup Z$ the maximum number of points that could be assigned to them is 36). Therefore, $|X \bigcup Y \bigcup Z|=19$ and each student in  $X \bigcup Y \bigcup Z$ is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade $A=25 - 19-6=0$, that is, no student gets A grade. Since there are $19-8=11$ students who are not weight lifters all these 11 students must be both swimmers and cyclists. (Similarly, there are 2 who are both swimmers and weight lifters and 6 who are both cyclists and weight lifters).

More later,

Nalin Pithwa

# Pre-RMO or RMO tutorial: Homi Bhabha Science Foundation

Some more questions based on logic only:

1. Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they made the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Nalin Pithwa.

# 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 . 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.

# RMO Training: more help from Nordic mathematical contest

Problem:

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.

Solution:

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. 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.

https://www.amazon.in/Nordic-Mathematical-Contest-1987-2009-Todev/dp/1450519830/ref=sr_1_1?s=books&ie=UTF8&qid=1518386661&sr=1-1&keywords=Nordic+mathematical+contest

# Journos have a problem!

A problem posed in Nordic Mathematical Contest 1987:

Question:

Nine journalists, each from a different country, participate in a press conference. None of them can speak more than three languages, and each two journalists have at least one common language. Prove that at least five of the journalists can speak the same language.

Solution1:

Assume the journalists are $J_{1}, J_{2}, J_{3}, \ldots, J_{9}$. Assume that no five of them have a common language. Assume the languages $J_{1}$ speaks are $L_{1}, L_{2}$ and $L_{3}$. Group $J_{2}, J_{3}, \ldots, J_{9}$ according to the language they speak with $J_{1}$. No group can have more than three members. So, either there are three groups of three members each, or two groups with three members and one with two. Consider the first alternative. We may assume that $J_{1}$ speaks $L_{1}$ with $J_{2}$, $J_{3}$, and $J_{4}$, $L_{2}$ with $J_{5}$, $J_{6}$, and $J_{7}$, and $L_{3}$ with $J_{8}$, $J_{9}$ and $J_{2}$. Now, $J_{2}$ speaks $L_{1}$ with $J_{1}$, $J_{3}$, and $J_{4}$, $L_{3}$ with $J_{1}$, $J_{8}$, and $J_{9}$. $J_{2}$ must speak a fourth language, $L_{4}$, with $J_{5}$, $J_{6}$, and $J_{7}$. But, now $J_{5}$ speaks both $L_{2}$ and $L_{4}$ with $J_{2}$, $J_{6}$, and $J_{7}$. So, $J_{5}$ has to use his third language with $J_{1}, J_{4}, J_{8}$ and $J_{9}$. This contradicts the assumption we made.

So, we now may assume that $J_{1}$ speaks $L_{3}$ only with $J_{8}$ and $J_{9}$. As $J_{1}$ is not special, we conclude that for each journalist $J_{k}$, the remaining eight are divided into three mutually exclusive language groups, one of which has only two members. Now, $J_{2}$ uses $L_{1}$ with three others, and there has to be another language he also speaks with three others. If this were $L_{2}$ or $L_{3}$, a group of five would arise (including $J_{1}$). So, $J_{2}$ speaks $L_{4}$ with three among $J_{5}, \ldots, J_{9}$. Either two of these three are among $J_{5}$, $J_{6}$, and $J_{7}$, or among $J_{8}, J_{9}$. Both alternatives lead to a contradiction to the already proved fact that no pair of journalists speaks two languages together. Hence, the proof. QED.

Reference:

Nordic Mathematical Contest, 1987-2009.