2019 AMC 10A Problem 11

Taking the prime factorization of $201^9,$ we get $3^9 \cdot 67^9.$ Note that a perfect square has even exponents for its prime factors, and a cube’s exponents are divisible by $3.$ There are $5$ options for an even exponent, from $0$ through $8,$ and $4$ options for multiples of $3,$ from $0$ through $9.$ This gives us $5^2$ options for the squares and $4^2$ options for the cubes. We have to subtract out the sixth powers, however. Using the same logic, sixth powers have to have exponents of prime factors be divisible by $6.$ There are $2$ options, $0$ and $6.$ This means that there are $2^2 = 4$ sixth powers. This gives us a total of $$25 + 16 - 4 = 37.$$ perfect squares or perfect cubes. Thus, C is the correct answer.