How many divisors of $99^{99}$ are perfect cubes?

Question

I think I managed to solve it by writing out the prime factors and then counting those whose exponents were multiples of 3. Am I correct?

Answer 1

The prime factorization of $99^{99}$ can be found as follows: $$99^{99}=(3^2*11)^{99}=3^{198}*11^{99}$$ Thus, any divisors of $99^{99}$ are in the form of $3^a\cdot 11^b$ where $a \leq 198$ and $b \leq 99$. For this to be a perfect cube, both $a$ and $b$ have to be multiples of $3$.

If $a$ is a multiple of $3$ and $a \leq 198$, that gives us: $$a \in \{3\cdot 0, 3\cdot 1, 3\cdot 2, ..., 3\cdot 66=198\}$$ Thus, there are $67$ possibilities for $a$.

If $b$ is a multiple of $3$ and $b \leq 99$, that gives us: $$b \in \{3\cdot 0, 3\cdot 1, 3\cdot 2, ..., 3\cdot 33=99\}$$ Thus, there are $34$ possibilities for $b$.

To get a perfect cube factor, we can choose any of these $a$ and choose any of these $b$, which are independent choices. Thus, from the Counting Principle, there are $67\cdot 34=2278$ (positive) factors of $99^{99}$ that are perfect cubes.

Answer 2

Since $99=3^2\cdot 11$, we have

$$99^{99}=3^{198}\cdot 11^{99}$$

so the divisors that are perfect cubes look like

$$\{1,3^3,3^6,\ldots,3^{198}\}\times\{1,11^3,11^6,\ldots,11^{99}\}$$