2025 AMC 8 Problems/Problem 22

Problem

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?

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution 1

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

Solution 2

Say Paulina placed $n$ coats. That will divide the 35 hooks into $n+1$ spaces and $35-n$ empty hooks. Therefore, $n+1|35-n.$ The values of $n$ that satisfy this are $n\in{1,2,3,5,8,11,17}$ The answer is $\boxed{\text{(D) }7}$ . ~Tacos_are_yummy_1

Solution 3 (Critical Thinking)

Say the number of empty coats before the first coat are $n$ .

By that, $n$ is the number of empty hooks between all coats. There is $n+1$ gaps in every situation because $n$ cannot be 0.

To solve the problem, you minus $2n$ from 35(The $n$ empty hook at the beginning and $n$ empty hook from the end). Then minus 1(The first occupied hook). After that, divide the rest of the number of hook by $n+1$ (The difference of hooks between coats). We do this because the first occupied hook is always after $n$ hooks, we just see if the rest can be divided by the difference.

To make it clear:

If $n$ = 1

1. 35 - 2n = 33

2. 33 - 1 = 32

3. 32/n+1 = 16

16 is a integer so this situation satisfies the conditions.

For $n$ = 2

We can skip to the third step since every time you add 1 to $n$ it two less then last situation. 30/n+1 = 10 This situation satisfies the conditions.

We keep doing this until the difference is greater than 35-(2n+1) and we see that 7 situations does satisfy the conditions. Note: This method is slow but is easy to check and understand. ~LIBAIKO

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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