2017 AMC 12B Problem 16

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}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.