IMO 1974 Problem 1

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 9m35s

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?

Exploration

The corrected argument begins by rechecking the structure of the process itself rather than inheriting any assumption from the flawed solution.

The game consists of three rounds, and each round assigns a permutation of the fixed multiset ${p,q,r}$ to $(A,B,C)$. Therefore each round independently uses $p$, $q$, and $r$ exactly once, but the same player may receive the same value in different rounds. This removes the incorrect restriction that a player can receive $r$ only once overall.

The total sum is $39$, so if there are $n$ rounds then $n(p+q+r)=39$. Since $p+q+r\ge 6$, one gets $n\le 6$. The divisors of $39$ are $1,3,13,39$, and only $n=3$ is compatible with $n\ge 2$. Hence $p+q+r=13$.

The remaining structure depends on finding integer triples $p<q<r$ summing to $13$ that allow three permutations to produce totals $A=20$, $B=10$, $C=9$, with the additional condition that in the last round $B$ receives $r$.

The feasible triples must be tested against the strong constraint that $A$ is significantly above average while $C$ is significantly below average, suggesting repeated large contributions to $A$ and repeated small contributions to $C$.

Exhausting admissible triples under $p+q+r=13$ and checking compatibility with $B$’s final condition isolates only a small number of candidates, which can then be tested against the round structure.

Problem Understanding

Three players receive values $p<q<r$ over three rounds, each round being a permutation of these three numbers. After three rounds the totals are fixed at $A=20$, $B=10$, $C=9$, and in the last round $B$ receives $r$. The task is to determine who received $q$ in the first round.

The problem reduces to determining all triples $(p,q,r)$ with $p+q+r=13$ that admit a decomposition into three permutations consistent with the totals and then identifying the forced placement of $q$ in the first round.

Key Observations

The number of rounds is exactly $3$, so each player receives exactly three values drawn from ${p,q,r}$.

The average per player is $13$, so deviations are $A=+7$, $B=-3$, $C=-4$. This indicates that $A$ must receive $r$ at least twice or receive one very large and two medium contributions, while $C$ must avoid $r$ and receive $p$ frequently.

The last round fixes $B$ at $r$, so $B$ has limited capacity to accumulate a large total, forcing strong structure in the first two rounds.

Enumerating all integer triples summing to $13$ reduces the search space to eight cases, and most are eliminated immediately by the requirement that $B$ achieves total $10$ using two remaining values from ${p,q,r}$ across the first two rounds.

Only two triples survive this constraint: $(1,4,8)$ and $(2,5,6)$.

Solution

From $n(p+q+r)=39$ and $p+q+r\ge 6$, one obtains $n\le 6$. Since $n$ divides $39$, the only possibility with $n\ge 2$ is $n=3$, hence $p+q+r=13$.

All ordered triples of positive integers $p<q<r$ with sum $13$ are

$$(1,2,10), (1,3,9), (1,4,8), (1,5,7), (2,3,8), (2,4,7), (2,5,6), (3,4,6).$$

The condition that in the last round $B$ receives $r$ implies that in the first two rounds $B$ receives two values from ${p,q,r}$ whose sum is $10-r$. Each such pair must be realizable as a sum of two elements from ${p,q,r}$.

Testing each triple against the requirement that $B$ attains total $10$ after three rounds with the last contribution fixed as $r$ eliminates all but $(1,4,8)$ and $(2,5,6)$.

For $(2,5,6)$, the last round gives $B=6$, so the first two rounds must give $B=4$. The only way to achieve $4$ using two values from ${2,5,6}$ is $2+2$, forcing $B$ to receive $2$ in both of the first two rounds. This leaves $(A,C)$ in the first two rounds alternating between $(5,6)$ and $(6,5)$. Any such arrangement yields first-two-round totals for $A$ equal to $10$, $11$, or $12$, depending on distribution, and none can reach the required $A$ total once the last round is imposed. This eliminates $(2,5,6)$.

For $(1,4,8)$, the last round gives $B=8$, so the first two rounds must give $B=2$, forcing $B$ to receive $1$ in both of the first two rounds. Thus in the first two rounds, $(A,C)$ must each receive ${4,8}$ once per round.

The last round splits $(p,q)=(1,4)$ between $A$ and $C$.

If $A=1$ and $C=4$ in the last round, then $A$ would need $19$ from two rounds using only ${4,8}$, which is impossible since the only achievable sums are $8$, $12$, and $16$.

If $A=4$ and $C=1$ in the last round, then $A$ must receive $16$ from the first two rounds using two values from ${4,8}$. This forces $A=8$ in both rounds and $C=4$ in both rounds, so both first two rounds are identical and equal to $(A,B,C)=(8,1,4)$.

In the first round, $q=4$ is received by $C$.

Thus the only consistent configuration forces $C$ to receive $q$ in the first round.

Verification of Key Steps

The restriction to $n=3$ follows from divisibility and the minimum value of $p+q+r$, which eliminates all other possibilities for the number of rounds.

The reduction to feasible triples is complete because all solutions of $p<q<r$ with sum $13$ are explicitly listed and checked against the constraint imposed by the last-round condition on $B$.

The elimination of $(2,5,6)$ is valid because forcing $B=2$ in both initial rounds leaves insufficient flexibility to reach $A=20$ and $C=9$ under permutation constraints, since the remaining values force incompatible totals for $A$.

The analysis of $(1,4,8)$ is exhaustive because both possible last-round assignments between $A$ and $C$ are tested, and only one is consistent with achievable two-round sums from ${4,8}$.

All remaining configurations are ruled out by direct contradiction with required partial sums, leaving a unique consistent structure.

Alternative Approaches

A matrix formulation encodes each round as a permutation matrix weighted by $(p,q,r)$ and converts the problem into solving a constrained integer linear system with permutation constraints. This makes the rigidity of the solution more transparent but requires systematic enumeration of feasible coefficient matrices.

Another approach uses deviation from the mean value $13$ to transform the problem into balancing positive and negative contributions over three rounds, then uses the last-round constraint to force a unique distribution of extremes, leading to the same elimination of all but one triple.