2017 AMC 10B Problem 20

Find the power of $2$ in $21!$ by counting multiples of $2$, $4$, $8$, $16$ up to $21$:
$$\left\lfloor \tfrac{21}{2} \right\rfloor + \left\lfloor \tfrac{21}{4} \right\rfloor + \left\lfloor \tfrac{21}{8} \right\rfloor + \left\lfloor \tfrac{21}{16} \right\rfloor = 10 + 5 + 2 + 1 = 18.$$
So $21! = 2^{18} \cdot M$ with $M$ odd.

A divisor of $21!$ is $2^k \cdot d$, where $0 \le k \le 18$ and $d$ is a divisor of $M$. For each of the $19$ possible values of $k$ there are the same number of choices of $d$, so the divisors split into $19$ equal groups according to their power of $2$. The odd divisors are exactly the group with $k = 0$.

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

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

The divisor count factors as a product over primes, $(e_2 + 1)(e_3 + 1)\cdots$, and "odd" is a condition on the exponent of $2$ alone. Whatever the other exponents do, the fraction of divisors with $2$-exponent zero is $\frac{1}{e_2 + 1}$. The huge number and the "over $60{,}000$" are pure distraction.

• Trying to use the given 60,000 divisor count or the decimal digits of 21!, instead of the exponent of 2.
• Miscounting the exponent of $2$ as $10$ (only the even numbers) or $21$, giving $\frac{1}{11}$ or $\frac{1}{22}$, or answering $\frac{1}{18}$ by forgetting the $+1$.