AMC 8 2025 Problem 22

AMC8 2025 年第 22 题

特殊数与不定方程★★★☆☆

A classroom has a row of 35 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 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina's pattern?

教室里有一排35个衣钩。Paulina希望衣服之间的间隔相等，使得第一件衣服前、最后一件衣服后以及每两件衣服之间的空衣钩数量相同。假设至少有1件衣服且至少有1个空衣钩。请问有多少种不同的衣服数量可以满足Paulina的要求？

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法·设两个未知数，加 1 凑因式分解

设挂了

c

件外套、每段空隙有

k

个空钩。先想清楚有几段空隙：

c

件外套像

c

根隔板，隔出「第一件前＋相邻两件之间（c-1 段）＋最后一件后」，共 c+1 段。题目要求每段一样多（

k

个），且

c\ge1

、

k\ge1

。

数全部 35 个钩子：外套占

c

个，空钩共

k(c+1)

个，

c+k(c+1)=35

展开得 kc+k+c=35。这种「两个未知数又相乘又相加」的方程有个标准招——两边加 1 凑乘积：

kc+k+c+1=36\quad\Rightarrow\quad(k+1)(c+1)=36

现在只需把 36 拆成两个因数的积。36 的因数有 1、2、3、4、6、9、12、18、36 共 9 个；因为

k\ge1

、

c\ge1

，两个因数都得

\ge2

，去掉带 1 的两种拆法 (1,36)、(36,1)，剩 9-2=7 种。每种拆法定出一个外套数：c+1=2,3,4,6,9,12,18，即 c=1,2,3,5,8,11,17。抽一个验算：c=17、k=1 时 17+1×18=35 ✓。共 7 种，选 D。

易错提醒：空隙是 c+1 段不是 c-1 段——第一件前、最后一件后都有空隙，题面写明了；另一个坑是丢掉「至少 1 件外套、至少 1 个空钩」，把带 1 的两种拆法也数进去得 9，正好是干扰项 E。

Solution 1

*(Elementary — two unknowns, then add 1 to factor.)*

Say

c

coats are hung with

k

empty hooks in every gap. First count the gaps:

c

coats act as

c

dividers, creating "before the first + between consecutive coats (c-1 gaps) + after the last," i.e. c+1 gaps, each holding

k

hooks, with

c\ge1

and

k\ge1

Counting all 35 hooks: coats take

c

, empty hooks total

k(c+1)

, so

c+k(c+1)=35

Expanding: kc+k+c=35. The standard move for "two unknowns multiplied and added" is to add 1 and factor:

kc+k+c+1=36\quad\Rightarrow\quad(k+1)(c+1)=36

Now just split 36 into two factors. Its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36 — nine of them. Since

k\ge1

and

c\ge1

, both factors must be

\ge2

, discarding (1,36) and (36,1): 9-2=7 ways. Each fixes one coat count: c+1=2,3,4,6,9,12,18, i.e. c=1,2,3,5,8,11,17. Spot-check c=17, k=1: 17+1×18=35 ✓. So 7 ways, choice D.

Watch out: there are c+1 gaps, not c-1 — the problem explicitly includes the gap before the first coat and after the last; and don't drop "at least 1 coat and at least 1 empty hook" — keeping the factor-1 splits gives 9, exactly trap choice E.

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