2025 AMC 8 第 22 题

先试着解答 2025 AMC 8 第 22 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 2025 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。

22.

一个教室有一排 3535 个挂衣钩。保利娜喜欢外套等距挂放,也就是说第一件外套前、最后一件外套后、以及每两件相邻外套之间的空钩数都相同。假设至少有 11 件外套,且至少有 11 个空钩。有多少种不同的外套数量能满足保利娜的排列方式?
A classroom has a row of 3535 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 11 coat and at least 11 empty hook. How many different numbers of coats can satisfy Paulina’s pattern?
22
44
55
77
99
答案:D
知识点:因数个数双射
小提示:
想象在最后额外增加一个挂有外套的钩子
Add one imaginary final coat to make equal blocks
大提示:
块大小和块数的乘积必须为 3636
The block size and number of blocks must multiply to 3636
视频讲解:
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文字解答:
想象在最后一件外套后再加一个挂着外套的钩子。现在共有 3636 个挂钩,并形成重复模式:每个块由若干空钩后接一件外套组成。设 bb 为每块中的位置数,设 dd 为块数。注意 dd 正好比原来的外套件数多一,因为末尾加了一件外套。
于是有 bd=36bd = 36。
对 bb 的限制是 b≥2b \geq 2,因为每块至少有一个空钩,并以一件外套结尾。
对 dd 的限制是 d≥2d \geq 2,因为原来至少有一件外套,末尾又额外加了一件。
所以只需数出有多少种方式把 3636 分解为两个至少为 22 的整数之积。
把 3636 分解成两个正整数之积的方式数,正好等于 3636 的因数个数。因为 36=22×3236 = 2^2 \times 3^2,所以 3636 的因数个数为 (2+1)(2+1)=9。 (2+1)(2+1) = 9\text{。}
其中正好两种不合格:1×361 \times 36 和 36×136 \times 1。所以答案是 9−2=79 - 2 = 7,选 D。
Imagine adding an extra coat hook with a coat on it after the last coat. Now, there will be 3636 coat hooks, and a repeating pattern, where each block of the pattern has a bunch of empty hooks followed by a coat. Let bb be the number of items in each block. Let dd be the number of blocks. Note that dd is exactly one more than the number of coats, because we added an extra coat at the end.
We then have bd=36.bd = 36.
The constraint on bb is that b≥2b \geq 2 because each block has at least one empty hook, and ends with a coat.
The constraint on dd is that d≥2d \geq 2 because there was at least one coat before, and we added one extra coat at the end.
So, we just need to find out how many ways there are to factorize 3636 into the product of two integers that are at least 2.2.
The number of ways to factorize 3636 into the product of two positive integers is exactly equal to the number of factors of 36.36. There is a formula for that: since 36=22×3236 = 2^2 \times 3^2, the number of factors of 3636 is (2+1)(2+1)=9. (2+1)(2+1) = 9.
Out of these factorizations, exactly two are disqualified: 1×361 \times 36 and 36×1.36 \times 1. So, the answer is 9−2=7,9 - 2 = 7, which is choice D.
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