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2023 AMC 12A 第 13 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

13.

在一场乒乓球锦标赛中，每位参赛者都与其他每位参赛者恰好比赛一次。虽然右手选手的人数是左手选手的两倍，但左手选手赢的场数比右手选手赢的场数多 40%40\%。（没有平局，也没有双利手选手。）总共进行了多少场比赛？

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was 40%40\% more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

1515

3636

4545

4848

6666

答案：B
知识点：组合整除性
难度评级：1660
小提示：

设有 LL 名左手选手和 2L2L 名右手选手，所以总人数为 3L3L。

Let there be LL left-handed and 2L2L right-handed players, so 3L3L players total

大提示：

若右手选手赢 RR 场，则总场数为 2.4R2.4R，所以总场数必须是 1212 的倍数

If right-handers win RR games, the total is 2.4R,2.4R, so the total must be a multiple of 1212

解答：

设有 LL 名左手选手和 2L2L 名右手选手，则共有 3L3L 人，总场数为 (3L2)\binom{3L}{2}。

若右手选手赢 RR 场，则左手选手赢 1.4R1.4R 场，所以总场数为 2.4R=125R2.4R=\tfrac{12}{5}R。要使它是整数场数，总场数必须是 1212 的倍数。

左手选手至多能赢下所有至少有一名左手选手参加的比赛，即 (L2)+2L2=5L2−L2\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2} 场。另一方面，他们必须赢 712(3L2)=7L(3L−1)8\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8 场。比较这两个量可得 21L−7≤20L−421L-7\le 20L-4，所以 L≤3L\le 3。

依次试 L=1,2,3L=1,2,3，总场数为 3,15,363,15,36；其中只有 3636 是 1212 的倍数，且可实现：33 名左手选手可以赢下全部 1818 场混合对局以及左手内部的 33 场，共 21=1.4⋅1521=1.4\cdot 15 场。

所以正确答案是 B。

Let there be LL left-handed and 2L2L right-handed players, for 3L3L players and (3L2)\binom{3L}{2} games total.

If right-handers win RR games, left-handers win 1.4R,1.4R, so the total is 2.4R=125R.2.4R=\tfrac{12}{5}R. For this to be an integer count, the total number of games must be a multiple of 12.12.

The left-handers can win at most every game involving at least one left-hander, namely (L2)+2L2=5L2−L2.\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}. On the other hand, they must win 712(3L2)=7L(3L−1)8\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8 games. Comparing these quantities gives 21L−7≤20L−4,21L-7\le 20L-4, so L≤3.L\le 3.

Testing L=1,2,3L=1,2,3 gives totals 3,15,36;3,15,36; only 3636 is a multiple of 12.12. It is achievable: the 33 left-handers can win all 1818 mixed games and their 33 internal games, giving 21=1.4⋅1521=1.4\cdot 15 wins.

Thus, the correct answer is B.

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