2017 AMC 12B Problems/Problem 16

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?

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

Solution

If a factor of $21!$ is odd, that means it contains no factors of $2$ . We can find the number of factors of two in $21!$ by counting the number multiples of $2$ , $4$ , $8$ , and $16$ that are less than or equal to $21$ .After some quick counting we find that this number is $10+5+2+1 = 18$ . If the prime factorization of $21!$ has $18$ factors of $2$ , there are $19$ choices for each divisor for how many factors of $2$ should be included ( $0$ to $18$ inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of $2$ is $0$ which is $\frac{1}{19} \Rightarrow \boxed{B}$ .

Solution by: vedadehhc

Page

Toolbox

Search

2017 AMC 12B Problems/Problem 16

Problem 16

Solution