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AMC10 2027 - Self-practice Set 3 Question 16

AMC10 · AMC10 2027 - Self-practice Set 3

Question 16

The number $21!=51\,090\,942\,171\,709\,440\,000$ has over $60\,000$ positive integer divisors. One is chosen uniformly at random. What is the probability that it is odd?
  1. $\frac1{21}$
  2. $\frac1{19}$
  3. $\frac1{18}$
  4. $\frac12$
  5. $\frac{11}{21}$