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2020 AMC 10A Problem 15

Problem 15 of 25IntermediateNumber TheoryCounting & Probability

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

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Solution

The prime factorization of 12!12! is 2103552711112^{10}3^5 5^2 7^1 11^1. Therefore 12!12! has (10+1)(5+1)(10+1)(5+1) (2+1)(1+1)(1+1)=792\cdot(2+1)(1+1)(1+1)=792 positive divisors. A square divisor must use only even exponents, giving 63211=366\cdot3\cdot2\cdot1\cdot1=36 square divisors. The probability is 36792=122\frac{36}{792}=\frac{1}{22}, so m+n=1+22=23m+n=1+22=23. Thus, E is the correct answer.

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Concepts: factor counting · perfect square · basic probability

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.