AMC8 2025

AMC8 2025 · Q22

AMC8 2025 · Q22. It mainly tests Arithmetic misc, GCD & LCM.

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?

一个教室有一排35个衣帽钩。Paulina喜欢外套等间距放置，使得第一个外套前、最后一个外套后以及每个外套与下一个外套之间都有相同数量的空钩。假设至少有1件外套且至少有1个空钩。有多少种不同的外套数量可以满足Paulina的模式？

(A) \ 2 \ 2

(B) \ 4 \ 4

(D) \ 7 \ 7

(E) \ 9 \ 9

Answer

Correct choice: (D)

正确答案：（D）

Solution

Suppose there are $c$ coats on the rack. Notice that there are $c+1$ "gaps" formed by these coats, each of which must have the same number of empty spaces (say, $k$). Then the values $k$ and $c$ must satisfy $c+k(c+1)=35 \implies kc+k+c=35$. We now use Simon's Favorite Factoring Trick as follows: \[kc+k+c=35\] \[\implies kc+k+c+1=36\] \[\implies (k+1)(c+1)=36\] Our only restrictions now are that $k>0 \implies k+1 > 1$ and $c>0 \implies c+1>1$. Other than that, each factor pair of $36$ produces a valid solution $(k,c)$, which in turn uniquely determines an arrangement. Since $36$ has $9$ factors, our answer is $9-2=\boxed{\textbf{(D)}~7}$. ~cxsmi

假设衣架上有$c$件外套。注意这些外套形成了$c+1$个“间隙”，每个间隙必须有相同数量的空位（设为$k$）。则$k$和$c$必须满足$c+k(c+1)=35 \implies kc+k+c=35$。我们现在使用Simon's Favorite Factoring Trick如下：\[kc+k+c=35\] \[\implies kc+k+c+1=36\] \[\implies (k+1)(c+1)=36\] 现在唯一的限制是$k>0 \implies k+1 > 1$且$c>0 \implies c+1>1$。除此之外，36的每个因子对产生一个有效的解$(k,c)$，从而唯一确定一种排列。由于36有9个因子，我们的答案是$9-2=\boxed{\textbf{(D)}~7}$。 ~cxsmi

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