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Source: 2017 AMC 10B #20 / AMC 12B #16

February 16, 2017
2017 AMC 10BAMC 10AMCprobability2017 AMC 12BAMC 12

Problem Statement

The number 21!=51,090,942,171,709,440,00021!=51,090,942,171,709,440,000 has over 60,00060,000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
<spanclass=′latex−bold′>(A)</span>121<spanclass=′latex−bold′>(B)</span>119<spanclass=′latex−bold′>(C)</span>118<spanclass=′latex−bold′>(D)</span>12<spanclass=′latex−bold′>(E)</span>1121<span class='latex-bold'>(A)</span> \frac{1}{21} \qquad <span class='latex-bold'>(B)</span> \frac{1}{19} \qquad <span class='latex-bold'>(C)</span> \frac{1}{18} \qquad <span class='latex-bold'>(D)</span> \frac{1}{2} \qquad <span class='latex-bold'>(E)</span> \frac{11}{21}
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