Pre-RMO or RMO sample problem solutions: logic questions

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

Amazon India link:

https://www.amazon.in/Problem-Primer-Olympiad-2Ed-Pranesachar/dp/8172862059/ref=sr_1_2?s=books&ie=UTF8&qid=1518891874&sr=1-2&keywords=problem+primer+for+the+olympiad

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 $1 \leq a < b <c \leq 6$ . 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).