Question

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?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

Solution:

Note that given any integer $z,$ we can represent it as the product as its even and odd components as $z=2^c d.$ This comes from the uniqueness of the prime factorization of integers, as we simply aggregate the odd and even primes (or in other words, $2,$ and everything else).

With this in mind, using the prime factorization of $21!,$ the even part of such a representation is $2^{18}.$ As such, let $21! = 2^{18}d,$ making $d$ an odd divisor of $21!.$

This means every odd divisor of $21!$ is a divisor of $d.$ For any odd divisor $x$ of $d,$ we know $2^kx$ is a divisor of $21!$ for $0 \leq k \leq 18.$ It is only odd for $k=0,$ and as such, there are $19$ possible values of $k.$

As such, the total probability is $\dfrac{1}{19}.$

Thus, the correct answer is B .

2017 AMC 10B Problem 20

Solution:

Problem 20 in Other Years