2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 15 Medium

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 mn, where m and n are relatively prime positive integers. What is m+n?

  • A.

    3

  • B.

    5

  • C.

    12

  • D.

    18

  • E.

    23

Answer:D

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. 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 !. Thus, there are 6 \cdot 3 \cdot 2 perfect squares. (For 2 , you can have 0,2,4,6,8, or 102 \text{~s}, etc.) The probability that the divisor chosen is a perfect square is \frac{6 \cdot 3 \cdot 2}{11 \cdot 6 \cdot 3 \cdot 2 \cdot 2}=\frac{1}{22} \Longrightarrow \frac{m}{n}=\frac{1}{22} \Longrightarrow m+n=1+22=\text { (E) } 23

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