Cube factors
How many factors of $9^9$ are perfect cubes?
How many factors of $9^9$ are perfect cubes?
This problem is taken from the World Mathematics Championships
Answer: 7
$9^9=\left(3^2\right)^9=3^{18}$
Factors of $3^{18}$: $1,3,3^2,3^3,...,3^{17},3^{18}$ (no others since $3$ is prime)
Cubes: The ones where the power is a multiple of $3$ are cubes, for example $3^{15} = 3^{3\times5} = \left(3^5\right)^3$
The ones where the power is not a multiple of $3$ are not cubes, because $3$ is prime, so there is no other way to divide the product into $3$ equal parts
Cube factors: $1, 3^3, 3^6, 3^9, 3^{12}, 3^{15}, 3^{18}$
There are $7$