AMC12 2017 B

Answer (B): There are $\left\lfloor \frac{21}{2}\right\rfloor+\left\lfloor \frac{21}{4}\right\rfloor+\left\lfloor \frac{21}{8}\right\rfloor+\left\lfloor \frac{21}{16}\right\rfloor=10+5+2+1=18$ powers of 2 in the prime factorization of $21!$. Thus $21!=2^{18}k$, where $k$ is odd. A divisor of $21!$ must be of the form $2^ib$ where $0\le i\le 18$ and $b$ is a divisor of $k$. For each choice of $b$, there is one odd divisor of $21!$ and 18 even divisors. Therefore the probability that a randomly chosen divisor is odd is $\frac{1}{19}$. In fact, $21!=2^{18}\cdot 3^9\cdot 5^4\cdot 7^3\cdot 11\cdot 13\cdot 17\cdot 19$, so it has $19\cdot 10\cdot 5\cdot 4\cdot 2\cdot 2\cdot 2\cdot 2=60{,}800$ positive integer divisors, of which $10\cdot 5\cdot 4\cdot 2\cdot 2\cdot 2\cdot 2=3{,}200$ are odd.