2017 AMC 10B Problems/Problem 20

Problem

The number $21!=51,090,942,171,709,440,000$ has over $60,000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}$

Solution

We note that the only thing that affects the parity of the factor are the powers of 2. There are $10+5+2+1 = 18$ factors of 2 in the number. Thus, there are 18 cases in which a factor of $21!$ would be even (have a factor of $2$ in its prime factorization), and 1 case in which a factor of $21!$ would be odd. Therefore, the answer is $\boxed{\textbf{(B)} \frac 1{19}}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2017 AMC 10B Problems/Problem 20

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