# 1974 IMO Problems/Problem 1

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## Problem

Three players $A, B$ and $C$ play the following game: On each of three cards an integer is written. These three numbers $p, q, r$ satisfy $0 < p < q < r$. The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.

This process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, $A$ has 20 counters in all, $B$ has 10 and $C$ has 9. At the last round $B$ received $r$ counters. Who received $q$ counters on the first round?

## Solution

Answer: player $C$.

Let $n$ be the number of rounds played, then obviously $n (p + q + r) = 20 + 10 + 9 = 39$. So $n$ must be a divisor of 39, i. e. $n \in \{ 1, 3, 13, 39 \}$. But $p \geq 1,$ $q \geq p + 1 \geq 2,$ $r \geq q + 1 \geq 3,$ so $p + q + r \geq 1 + 2 + 3 = 6$ and $n = \frac{39}{p + q + r} \leq \frac{39}{6} < 7$. Also, by condition $n \geq 2,$ so we conclude to $n = 3$ and $p + q + r = \frac{39}{n} = 13$.

As $B$ received 10 counters including $r$ counters at the last round, $10 \geq p + p + r \geq 2 + r,$ so $r \leq 8$. On the other hand, from number of counters received by $A$ we get $r > \frac{20}{3} > 6$. So $r \in \{ 7, 8 \}$.

If $r = 7,$ from number of counters received by $B$ we get $3 = 10 - r \in \{ 2p, p + q, 2q \},$ and since 3 is odd, we get $p + q = 3$. But then $p + q + r = 3 + 7 = 10 \neq 13$ - a contradiction.

So $r = 8$ and $2 = 10 - r \in \{ 2p, p + q, 2q \}$. On the other hand, $2 \leq 2p < p + q < 2q,$ so $2 = 2p,$ $p = 1$ and $q = 13 - p - r = 4$. It is easy to check that the only way to distribute the counters for players is $20 = r + r + q,$ $10 = p + p + r$ and $9 = q + q + p,$ so player $C$ received $q$ counters on the first round.