Question

一个有 $12$ 支队伍的联赛进行循环赛，每支队伍与其他每支队伍恰好比赛一次。比赛要么一队获胜，要么以平局结束。每赢一场得 $2$ 分，每平一场得 $1$ 分。关于这 $12$ 个得分组成的列表，下列哪一项不一定为真？

A league with $12$ teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores $2$ points for every game it wins and $1$ point for every game it draws. Which of the following is not a true statement about the list of $12$ scores?

奇数得分的个数一定是偶数。

There must be an even number of odd scores.

偶数得分的个数一定是偶数。

There must be an even number of even scores.

不可能有两个得分为 $0$ 的队伍。

There cannot be two scores of $0.$

得分总和至少为 $100$ 。

The sum of the scores must be at least $100.$

最高得分至少为 $12$ 。

The highest score must be at least $12.$

答案：E

知识点：奇偶性不变量反例

难度评级：1660

小提示：

每场比赛总共恰好贡献 $2$ 分，所以所有得分的总和是固定的

Every game contributes exactly $2$ points in total, so the sum of all scores is fixed

大提示：

考虑如果每一场比赛都以平局结束会怎样

Consider what happens if every single game ends in a draw

解答：

每支队伍都打 $11$ 场，因此 $12$ 支队伍共进行 $\dfrac{12\cdot 11}{2} = 66$ 场比赛。每场比赛给得分列表增加 $2$ 分，所以所有得分的总和为 $66\cdot 2 = 132$ 。

如果每场比赛都是平局，每支队伍得 $11$ 分，所以最高得分不必达到 $12$ ；因此命题 $\text{(E)}$ 可能不成立。其他命题总成立：总和 $132 \ge 100$ ；总和为偶数，迫使奇数得分的个数为偶数，因而偶数得分的个数也为偶数；两支队伍不可能都得 $0$ 分，因为它们之间的比赛至少会给其中一队一分。

因此，正确答案是 E。

Each of the $12$ teams plays $11$ games, so $\dfrac{12\cdot 11}{2} = 66$ games are played, and each game adds $2$ points to the list. The total of all scores is $66\cdot 2 = 132.$

If every game is a draw, each team scores $11,$ so the highest score need not reach $12;$ thus statement $\text{(E)}$ can fail. The other statements always hold: the sum $132 \ge 100,$ the sum being even forces an even number of odd scores and hence an even number of even scores, and two teams cannot both score $0$ because their mutual game gives at least one of them a point.

Thus, the correct answer is E.

2015 AMC 12A 第 13 题

答案：E

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 13 题