AMC8 2015

AMC8 2015 · Q24

AMC8 2015 · Q24. It mainly tests Linear equations, Inequalities with integers (floor/ceiling basics).

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?

一个棒球联盟由两个四队分组成。每个队与其分内其他队各打N场比赛。每个队与另一分的所有队各打M场比赛，其中N > 2M且M > 4。每个队打76场比赛的赛程。一个队在其本分内打多少场比赛？

(A) 36 36

(B) 48 48

(D) 60 60

(E) 72 72

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The number of games played by a team is $3N + 4M = 76$. Because $M > 4$ and $N > 2M$ it follows that $N > 8$. Because $4$ divides both $76$ and $4M$, $4$ must divide $3N$ and hence $N$. If $N = 12$ then $M = 10$ and the condition $N > 2M$ is not satisfied. If $N \ge 20$ then $M \le 4$ and the condition $M > 4$ is not satisfied. So the only possibility is $N = 16$ and $M = 7$. So each team plays $3 \cdot 16 = 48$ games within its division and $4 \cdot 7 = 28$ games against the other division.

答案（B）：一支队伍比赛的场数为 $3N + 4M = 76$。因为 $M > 4$ 且 $N > 2M$，可得 $N > 8$。由于 $4$ 同时整除 $76$ 和 $4M$，所以 $4$ 必须整除 $3N$，从而也必须整除 $N$。如果 $N = 12$，则 $M = 10$，但不满足条件 $N > 2M$。如果 $N \ge 20$，则 $M \le 4$，但不满足条件 $M > 4$。因此唯一可能是 $N = 16$ 且 $M = 7$。所以每支队伍在本分区内打 $3 \cdot 16 = 48$ 场比赛，并与另一个分区打 $4 \cdot 7 = 28$ 场比赛。

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