Three players play the following game. There are three cards each with a different positive integer. In each round the cards are randomly dealt to the players and each receives the number of counters on his card. After two or more rounds, one player has received 20, another 10 and the third 9 counters. In the last round the player with 10 received the largest number of counters. Who received the middle number on the first round?
Solution
The player with 9 counters.
The total of the scores, 39, must equal the number of rounds times the total of the cards. But 39 has no factors except 1, 3, 13 and 39, the total of the cards must be at least 1 + 2 + 3 = 6, and the number of rounds is at least 2. Hence there were 3 rounds and the cards total 13.
The highest score was 20, so the highest card is at least 7. The score of 10 included at least one highest card, so the highest card is at most 8. The lowest card is at most 2, because if it was higher then the highest card would be at most 13 - 3 - 4 = 6, whereas we know it is at least 7. Thus the possibilities for the cards are: 2, 3, 8; 2, 4, 7; 1, 4, 8; 1, 5, 7. But the only one of these that allows a score of 20 is 1, 4, 8. Thus the scores were made up: 8 + 8 + 4 = 20, 8 + 1 + 1 = 10, 4 + 4 + 1 = 9. The last round must have been 4 to the player with 20, 8 to the player with 10 and 1 to the player with 9. Hence on each of the other two rounds the cards must have been 8 to the player with 20, 1 to the player with 10 and 4 to the player with 9.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
7 Oct 1998