2020 AMC 10A Problem 15

Below is the video solution and professionally curated solution for Problem 15 of the 2020 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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Concepts:factor counting perfect square basic probability

Difficulty rating: 1420

15.

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 $m+n?$

$3$

$5$

$12$

$18$

$23$

Video solution:

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Written solution:

The prime factorization of $12!$ is $2^{10}3^5 5^2 7^1 11^1$ . Therefore $12!$ has $(10+1)(5+1)(2+1)(1+1)(1+1)=792$ positive divisors.

A square divisor must use only even exponents, giving $6\cdot3\cdot2\cdot1\cdot1=36$ square divisors. The probability is $36/792=1/22$ , so $m+n=1+22=23$ . Thus, E is the correct answer.

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