1974 IMO Problems/Problem 1
Problem
Three players and play the following game: On each of three cards an integer is written. These three numbers satisfy . 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, has 20 counters in all, has 10 and has 9. At the last round received counters. Who received counters on the first round?
Solution
Answer: player .
Let be the number of rounds played, then obviously . So must be a divisor of 39, i. e. . But so and . Also, by condition so we conclude to and .
As received 10 counters including counters at the last round, so . On the other hand, from number of counters received by we get . So .
If from number of counters received by we get and since 3 is odd, we get . But then - a contradiction.
So and . On the other hand, so and . It is easy to check that the only way to distribute the counters for players is and so player received counters on the first round.
See also
1974 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |