Question

一个棒球联盟由两个四队分区组成。每支球队与同分区其他每支球队比赛 $N$ 场。每支球队与另一个分区的每支球队比赛 $M$ 场，其中 $N > 2M$ ，且 $M > 4$ 。每支球队赛程共 $76$ 场。

每支球队在本分区内打多少场比赛？

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N > 2M$ and $M > 4.$ Each team plays a $76$ game schedule.

How many games does a team play within its own division?

$36$

$48$

$54$

$60$

$72$

答案：B

知识点：丢番图方程模运算

难度评级：1560

视频讲解：

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文字解答：

每支球队在本分区内打 $3N$ 场，对另一个分区打 $4M$ 场，所以 $3N+4M=76.$

因为 $N>2M$ ，所以 $3N>6M$ ，从而 $76=3N+4M>10M$ 。因此 $M<7.6$ 。又 $M>4$ ，所以 $M\in\{5,6,7\}$ 。

将 $3N+4M=76$ 模 $3$ ，得 $M\equiv1\pmod3$ ，所以 $M=7$ 。因此每支球队打 $4M=28$ 场非分区比赛，本分区比赛为 $76-28=48$ 场。

所以正确答案是 B。

Each team plays $3N$ games within its own division and $4M$ games against the other division, so $3N+4M=76.$

Since $N>2M$ , we have $3N>6M$ , and hence $76=3N+4M>10M$ . Thus $M<7.6$ . Together with $M>4$ , this gives $M\in\{5,6,7\}$ .

Reducing $3N+4M=76$ modulo $3$ gives $M\equiv1\pmod3$ , so $M=7$ . Therefore the team plays $4M=28$ non-division games and $76-28=48$ division games.

Thus, B is the correct answer.

2015 AMC 8 第 24 题

答案：B

难度评级：1560

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