AMC 8 2015 Problem 24
AMC8 2015 年第 24 题
特殊数与不定方程★★★★☆
A baseball league consists of two four-team divisions. Each team plays every other team in its division games. Each team plays every team in the other division 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?
一个棒球联盟有两组组成，每组 4 支队伍。每组里面的每支队伍和自己组内的其他每支队伍都打 N 场比赛。每支队伍和另一组的每支队伍都打 M 场比赛，且 N>2M，M>4，每支队伍共需要打 76 场比赛，那么每支队伍和自己组内的队伍共需要打多少场比赛？
- A.36
- B.48
- C.54
- D.60
- E.72
正确答案 · Correct Answer
B
解析 · Solution
解法1
常规解法 · 缩范围试值
由 N>2M 得 3N>6M，代入 3N+4M=76：。又 M>4，故 。逐个试 3N=76-4M 是否被 3 整除：（否），（否），（是），验 16>14 ✓。队内 3N=48。
巧解 · 取模 3
3N=76-4M 左边 ；、，故 。在 内唯一解 M=7，立得 N=16、队内 48。一步锁定，免试值。
Solution 1
Elementary · narrow the range, then test
On one team there are games within the division and across, so 3N+4M=76. From N>2M we get 3N>6M, hence 76-4M>6M, i.e. M<7.6; with M>4, . Test whether 3N=76-4M is divisible by 3: (no), (no), (yes), giving N=16 (and 16>14 ✓). So 3N=48 games within the division.
Slick · work mod 3
In 3N=76-4M the left side is ; since and , we need . The only such with is M=7, giving N=16 and 3N=48 — pinned in one step, no trial needed.
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