Difference between revisions of "2025 AMC 8 Problems/Problem 22"

Revision as of 20:53, 29 January 2025

Problem 22

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: $kc+k+c=35$ $kc+k+c+1=36$ $(k+1)(c+1)$ 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

@@ Line 1: / Line 1: @@
-The 2025 AMC 8 is not held yet. Please do not post false problems.
+==Problem 22==
+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?
+<math>\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9</math>
+==Solution 1==
+Suppose there are <math>c</math> coats on the rack. Notice that there are <math>c+1</math> "gaps" formed by these coats, each of which must have the same number of empty spaces (say, <math>k</math>). Then the values <math>k</math> and <math>c</math> must satisfy <math>c+k(c+1)=35 \implies kc+k+c=35</math>. We now use Simon's Favorite Factoring Trick as follows: <cmath>kc+k+c=35</cmath> <cmath>kc+k+c+1=36</cmath> <cmath>(k+1)(c+1)</cmath> Our only restrictions now are that <math>k>0 \implies k+1 > 1</math> and <math>c>0 \implies c+1>1</math>. Other than that, each factor pair of <math>36</math> produces a valid solution <math>(k,c)</math>, which in turn uniquely determines an arrangement. Since <math>36</math> has <math>9</math> factors, our answer is <math>9-2=\boxed{\textbf{(D)}~7}</math>. ~cxsmi

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Difference between revisions of "2025 AMC 8 Problems/Problem 22"

Revision as of 20:53, 29 January 2025

Problem 22

Solution 1