Question

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 $40\%$ 。（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 $40\%$ 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?

$15$

$36$

$45$

$48$

$66$

答案：B

知识点：组合整除性

难度评级：1660

解答：

设有 $L$ 名左手选手和 $2L$ 名右手选手，则共有 $3L$ 人，总场数为 $\binom{3L}{2}$ 。

若右手选手赢 $R$ 场，则左手选手赢 $1.4R$ 场，所以总场数为 $2.4R=\tfrac{12}{5}R$ 。 $12$ 。

要成为整数场数，总场数必须是 $\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}$ 的倍数。 $\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8$ $21L-7\le 20L-4$ ， $L\le 3$ 。

按选项或依次试 $L=1,2,3$ ，总场数为 $3,15,36$ ；其中只有 $36$ 是 $12$ 的倍数，且可实现： $3$ 名左手选手可以赢下全部 $18$ 场混合对局以及左手内部的 $3$ 场，共 $21=1.4\cdot 15$ 场。

所以正确答案是 B。

Let there be $L$ left-handed and $2L$ right-handed players, for $3L$ players and $\binom{3L}{2}$ games total.

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

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

Testing $L=1,2,3$ gives totals $3,15,36;$ only $36$ is a multiple of $12.$ It is achievable: the $3$ left-handers can win all $18$ mixed games and their $3$ internal games, giving $21=1.4\cdot 15$ wins.

Thus, the correct answer is B.

2023 AMC 12A 第 13 题

答案：B

难度评级：1660

解答：

其他年份的第 13 题