AMC12 2015 A

AMC12 2015 A · Q13

AMC12 2015 A · Q13. It mainly tests Arithmetic misc, Basic counting (rules of product/sum).

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?

一个有 12 个队的联赛举行循环赛，每队与其他每队恰好比赛一次。比赛要么一方获胜，要么平局。一队每胜一场得 2 分，每平一场得 1 分。关于 12 个得分列表，下列哪个不是真命题？

(A) There must be an even number of odd scores. 奇数比分的个数必须是偶数。

(B) There must be an even number of even scores. 偶数比分的个数必须是偶数。

(D) The sum of the scores must be at least 100. 比分总和至少为100。

(E) The highest score must be at least 12. 最高分至少为12。

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Note that each of the 12 teams plays 11 games, so $\frac{12\cdot 11}{2}=66$ games are played in all. If every game ends in a draw, then each team will have a score of 11, so statement (E) is not true. Each of the other statements is true. Each of the games generates 2 points in the score list, regardless of its outcome, so the sum of the scores must be $66\cdot 2=132$; thus (D) is true. Because the sum of an odd number of odd numbers plus any number of even numbers is odd, and 132 is even, there must be an even number of odd scores; thus (A) is true. Because there are 12 scores in all, there must also be an even number of even scores; thus (B) is true. Two teams cannot both have a score of 0 because the game between them must result in 1 point for each of them or 2 points for one of them; thus (C) is true.

答案（E）：注意这12支队每支要打11场比赛，所以总比赛场数为$\frac{12\cdot 11}{2}=66$。如果每场比赛都以平局结束，那么每支队的得分都是11，因此陈述（E）不成立。其他各条陈述都成立。每场比赛无论结果如何，在得分表中都会产生2分，因此总得分之和必为$66\cdot 2=132$；所以（D）成立。因为“奇数个奇数之和再加上任意多个偶数之和”必为奇数，而132是偶数，所以奇数得分的个数必须为偶数；因此（A）成立。因为一共有12个得分，所以偶数得分的个数也必须为偶数；因此（B）成立。两支队不可能同时得分为0，因为它们之间的比赛结果必然是双方各得1分（平局）或其中一方得2分；因此（C）成立。

Topics