2020 AMC 10A Problems/Problem 15

Problem

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$

Solution

The prime factorization of $12!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$ . This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that $7$ and $11$ can not be in the prime factorization of a perfect square because there is only one of each in $12!.$ Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$ , you can have $0$ , $2$ , $4$ , $6$ , $8$ , or $1$ 0 $2$ s, etc.) The probability that the divisor chosen is a perfect square is $\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m + n = 1 + 22 = 23 \implies \boxed{\textbf{(E) } 23 }$

~mshell214

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2020 AMC 10A Problems/Problem 15

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