PUMaC 2010 · 组合（B 组） · 第 5 题

PUMaC 2010 — Combinatorics (Division B) — Problem 5

3 n people take part in a chess tournament: n girls and 2 n boys. Each participant plays with each of the others exactly once. There were no ties and the number of games won by the girls 7 is the number of games won by the boys. How many people took part in the tournament? 5

解析

3 n people take part in a chess tournament: n girls and 2 n boys. Each participant plays with each of the others exactly once. There were no ties and the number of games won by the girls is 7 / 5 the number of games won by the boys. How many people took part in the tournament? 2 Answer: 9 ( ) 3 n 7 Solution: The number of games won by the girls is 7 / 12 the total number of games, or . 12 2 ( ) n The girls must win at least games, since a girl must win any game between two girls. The 2 ( ) ( ) 3 n 2 n girls can win at most − games, since a boy must win any game between two boys. 2 2 Therefore, we have ( ) ( ) ( ) ( ) n 7 3 n 3 n 2 n ≤ ≤ − 2 12 2 2 2 n ( n − 1) 7 3 n (3 n − 1) 3 n (3 n − 1) 2 n (2 n − 1) ≤ · ≤ − 2 12 2 2 2 4 n ( n − 1) ≤ 7 n (3 n − 1) ≤ 12 n (3 n − 1) − 8 n (2 n − 1) 2 2 2 4 n − 4 n ≤ 21 n − 7 n ≤ 20 n − 4 n 2 2 0 ≤ 17 n − 3 n ≤ 16 n 2 The first part of this inequality holds for all n , but the second part implies that n ≤ 3 n , 3 n (3 n − 1) so n ≤ 3. We also know that the total number of games, is divisible by 12. Then 2 3 n (3 n − 1) is divisible by 24, so n (3 n − 1) is divisible by 8. Since n and 3 n − 1 have different parity, one of n, 3 n − 1 is divisible by 8. Therefore, n = 3, so that 3 n = 9.