2015 AMC 8 第 24 题

先试着解答 2015 AMC 8 第 24 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2015 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

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 NN 场。每支球队与另一个分区的每支球队比赛 MM 场,其中 N>2MN > 2M,且 M>4M > 4。每支球队赛程共 7676 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 NN games. Each team plays every team in the other division MM games with N>2MN > 2M and M>4.M > 4. Each team plays a 7676 game schedule.
How many games does a team play within its own division?
3636
4848
5454
6060
7272
答案:B
知识点:丢番图方程模运算
小提示:
写出赛程方程 3N+4M=763N+4M=76
Write the schedule equation 3N+4M=76.3N+4M=76.
大提示:
使用 N>2MN>2M、M>4M>4,以及方程模 33 的结果
Use N>2M,N>2M, M>4,M>4, and the equation modulo 3.3.
视频讲解:
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文字解答:
每支球队在本分区内打 3N3N 场,对另一个分区打 4M4M 场,所以 3N+4M=76。3N+4M=76\text{。}
因为 N>2MN>2M,所以 3N>6M3N>6M,从而 76=3N+4M>10M76=3N+4M>10M。因此 M<7.6M<7.6。又 M>4M>4,所以 M∈{5,6,7}M\in\{5,6,7\}。
将 3N+4M=763N+4M=76 模 33,得 M≡1(mod3)M\equiv1\pmod3,所以 M=7M=7。因此每支球队打 4M=284M=28 场非分区比赛,本分区比赛为 76−28=4876-28=48 场。
所以正确答案是 B。
Each team plays 3N3N games within its own division and 4M4M games against the other division, so 3N+4M=76.3N+4M=76.
Since N>2M,N>2M, we have 3N>6M,3N>6M, and hence 76=3N+4M>10M.76=3N+4M>10M. Thus M<7.6.M<7.6. Together with M>4,M>4, this gives M∈{5,6,7}.M\in\{5,6,7\}.
Reducing 3N+4M=763N+4M=76 modulo 33 gives M≡1(mod3),M\equiv1\pmod3, so M=7.M=7. Therefore the team plays 4M=284M=28 non-division games and 76−28=4876-28=48 division games.
Thus, B is the correct answer.
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