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?

A. 36

B. 48

C. 54

D. 60

E. 72

1. 2015 AMC 8 官方解析

Solution 1：每支队伍在组内打 $3N$ 场、跨组打 $4M$ 场，$3N + 4M = 76$。因为 $M > 4$，从 $M = 5$ 试起：$M = 5$ 不行（$76 - 20 = 56$ 不是 3 的倍数）；$M = 6$ 不行（$52$ 不是 3 的倍数）；$M = 7$ 可以，得 $N = 16$，于是组内共 $3 \times 16 = 48$ 场。$M = 10$ 看似可行，但那样 $N = 12$，不满足 $N > 2M$。

Solution 2：$76 = 3N + 4M > 10M$，得 $M \leqslant 7$。又 $M > 4$，故 $M = 5, 6, 7$。因为 $4M \equiv 1 \pmod 3$，必须 $M \equiv 1 \pmod 3$，所以 $M = 7$，得 $3N = 48$。