2017 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2017 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 12B solutions, or check the answer key.

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Concepts:Legendre’s Formula factor counting prime factorization

Difficulty rating: 1730

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?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

Solution:

The exponent of $2$ in $21!$ is $\lfloor 21/2 \rfloor + \lfloor 21/4 \rfloor + \lfloor 21/8 \rfloor + \lfloor 21/16 \rfloor = 10 + 5 + 2 + 1 = 18.$ Every divisor has the form $2^i b$ with $0 \le i \le 18$ and $b$ odd; it is odd exactly when $i = 0.$ So the fraction of odd divisors is $\dfrac{1}{18 + 1} = \dfrac{1}{19}.$

Thus, the correct answer is B.

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