# 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 1

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 $\boxed{\textbf{(B)} \frac 1{19}}$.

Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21! into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.

## Solution 2 (Constructive Counting)

Consider how to construct any divisor $D$ of $21!$. First by Legendre's theorem for the divisors of a factorial (see here: Legendre's Formula), we have that there are a total of 18 factors of 2 in the number. $D$ can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for $D$ to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases = $\boxed{\textbf{(B)} \frac 1{19}}$

## Solution 3

We can find the prime factorization of 21!. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us $2^{18}$ x $3^{9}$ x $5^{4}$ x $7^{3}$ x 11 x 13 x 17 x 19. To find the number of odd divisors, we pretend as if the $2^{18}$ doesn't exist and count the other divisors: 10 x 5 x 4 x 2 x 2 x 2 x 2. The total number of divisors are 19 x 10 x 5 x 4 x 2 x 2 x 2 x 2. Dividing gives (B) 1/19.

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