2020 AMC 10A Problem 15

Question

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?

3

5

12

18

23

Factor

12!

into primes

A square divisor must choose only even prime exponents

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The prime factorization of

12!

is

2^{10}3^5 5^2 7^1 11^1

. Therefore

12!

has

(10+1)(5+1)

\cdot(2+1)(1+1)(1+1)=792

positive divisors.

A square divisor must use only even exponents, giving

6\cdot3\cdot2\cdot1\cdot1=36

square divisors. The probability is

\frac{36}{792}=\frac{1}{22}

, so

m+n=1+22=23

. Thus, E is the correct answer.

Problem 15 in other years