2020 AMC 10A Problem 15

First factor $12!$. The exponent of $2$ is $\lfloor 12/2 \rfloor + \lfloor 12/4 \rfloor + \lfloor 12/8 \rfloor = 6 + 3 + 1 = 10$; of $3$ it is $4 + 1 = 5$; of $5$ it is $2$; and $7$, $11$ appear once each:
$$12! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11.$$

A divisor picks an exponent for each prime from $0$ up to the exponent in $12!$, so the number of divisors is
$$11 \cdot 6 \cdot 3 \cdot 2 \cdot 2 = 792.$$

A divisor is a perfect square exactly when every exponent it picks is even. The even choices are $0,2,\dots,10$ for $2$ (six), $0,2,4$ for $3$ (three), $0,2$ for $5$ (two), and only $0$ for $7$ and $11$:
$$6 \cdot 3 \cdot 2 \cdot 1 \cdot 1 = 36.$$

The probability is $\frac{36}{792} = \frac{1}{22}$, so $m + n = 1 + 22 = 23$.

The answer is $\boxed{\textbf{(E)}\ 23}$.

Divisors of $N = \prod p_i^{e_i}$ correspond to exponent tuples, so any property defined exponent-by-exponent (being a square, being coprime to something) is counted by a product of per-prime counts. The probability question is just a ratio of two such products.

• Miscounting the exponent of 2 in 12! (it is 6 + 3 + 1 = 10) or forgetting that exponent 0 counts as even for square divisors.
• Reporting $m + n$ for the unreduced fraction $\frac{36}{792}$, or answering the probability's denominator alone.