Cube factors
How many factors of $9^9$ are perfect cubes?
Age
14 to 16
Challenge level
Problem
How many factors of $9^9$ are perfect cubes?
This problem is taken from the World Mathematics Championships
Student Solutions
Answer: 7
$9^9=\left(3^2\right)^9=3^{18}$
The factors of $3^{18}$ are $1,3,3^2,3^3,...,3^{17},3^{18}$. (There are no others since $3$ is prime.)
Cubes: The factors where the power is a multiple of $3$ are cubes, for example $3^{15} = 3^{3\times5} = \left(3^5\right)^3$.
The factors 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.
This means the cube factors are $1, 3^3, 3^6, 3^9, 3^{12}, 3^{15}, 3^{18}$.
Therefore there are $7$ cube factors of $9^9$.