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. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. 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!.) The probability that the divisor chosen is a perfect square is 1/22. m + n = 23 $\implies \boxed{\textbf{(E) } 23 }$

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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