Question

数 $21!$ $= 51{,}090{,}942{,}171{,}709{,}440{,}000$ 有超过 $60{,}000$ 个正整数因数。从这些因数中随机选一个。它是奇数的概率是多少？

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?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

答案：B

知识点：勒让德公式因数个数质因数分解

难度评级：1730

解答：

$21!$ 中 $2$ 的指数为 $\lfloor 21/2 \rfloor$ $+ \lfloor 21/4 \rfloor$ $+ \lfloor 21/8 \rfloor$ $+ \lfloor 21/16 \rfloor$ $= 10 + 5 + 2 + 1$ $= 18$ 。每个因数都形如 $2^i b$ ，其中 $0 \le i \le 18$ 且 $b$ 为奇数；它为奇数正好在 $i = 0$ 时。因此奇因数的比例为 $\dfrac{1}{18 + 1} = \dfrac{1}{19}$ 。

所以正确答案是 B。

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.

2017 AMC 12B 第 16 题

答案：B

难度评级：1730

解答：

其他年份的第 16 题