2017 AMC 10B Problem 20

Try a small case first. Take $12 = 2^2 \cdot 3$. Its divisors, sorted by how many $2$s they contain:

Each group has the same size, because each group is just the odd divisors $1, 3$ multiplied by $1$, $2$ or $4$. The odd divisors are one group out of $3$. The same idea works for $21!$; we only need to know how many $2$s it contains.

1. Count the $2$s in $21! = 1 \cdot 2 \cdot 3 \cdots 21$.
- Every even number gives at least one $2$: there are $10$ even numbers up to $21$.
- Multiples of $4$ give a second $2$: $4, 8, 12, 16, 20$, that is $5$ more.
- Multiples of $8$ give a third: $8, 16$, that is $2$ more.
- Multiples of $16$ give a fourth: $16$, that is $1$ more.
- Total: $10 + 5 + 2 + 1 = 18$.
2. So $21! = 2^{18} \cdot M$, where $M$ is odd.
3. Every divisor of $21!$ is (a power of $2$ from $2^0$ to $2^{18}$) times (a divisor of $M$). That is $19$ possible powers of $2$, and each one goes with the same list of divisors of $M$. So the divisors split into $19$ groups of equal size.
4. The odd divisors are exactly the group with $2^0 = 1$, one group out of $19$.

The probability is $\frac{1}{19}$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{19}}$.

"Odd" depends only on the number of $2$s in a divisor, and every allowed number of $2$s (from $0$ up to $18$) comes with the same set of choices for the other primes. So the fraction of odd divisors is always $1$ divided by (the number of $2$s plus one). The huge number and the "over $60{,}000$" are there only to distract.