2020 AMC 10A Problem 15

Attempt Problem 15 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution 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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15.

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?

33

55

1212

1818

2323

Answer: E
Concepts:factor countingperfect squarebasic probability
Difficulty rating: 1420
Video solution:
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Written 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 36/792=1/2236/792=1/22, so m+n=1+22=23m+n=1+22=23. Thus, E is the correct answer.

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