2019 AMC 10A Problem 11

1. Factor the base. $201 = 3 \cdot 67$, and both $3$ and $67$ are prime. So $201^9 = 3^9 \cdot 67^9$.
2. What does a divisor look like? Every divisor is $3^a \cdot 67^b$, where $a$ can be any of $0, 1, \ldots, 9$ and $b$ can be any of $0, 1, \ldots, 9$.
3. When is it a perfect square? A number is a perfect square when every prime in it appears an even number of times. For example, $3^4 \cdot 67^2 = (3^2 \cdot 67)^2$. So $a$ and $b$ must both be even: $0, 2, 4, 6, 8$. That is $5$ choices for $a$ and $5$ for $b$, so $5 \cdot 5 = 25$ squares.
4. When is it a perfect cube? Each exponent must be a multiple of $3$: $0, 3, 6, 9$. That is $4 \cdot 4 = 16$ cubes.
5. Which are both? The exponents must be even and multiples of $3$, so multiples of $6$: $0$ or $6$. That gives $2 \cdot 2 = 4$ divisors, and they were counted in both step 3 and step 4.
6. Count each divisor once. Add the two lists and remove the ones counted twice:
$$25 + 16 - 4 = 37.$$

The answer is $\boxed{\textbf{(C)}\ 37}$.

Once a number is broken into primes, its divisors are just choices of exponents, so "how many divisors are squares" becomes "how many ways to pick even exponents". For "this or that (or both)", add the two counts and subtract the overlap, since those were counted twice. And always check whether the base is really prime: $201$ is not.