How many three-digit perfect cubes are divisible by $9?$
A cube is divisible by $9$ if the number being cubed is divisible by $\sqrt[3]{9}=3^{\frac{2}{3}}.$ Since a perfect cube is the cube of an integer, we need the number being cubed to be a multiple of $3,$ so the cube is of the form $(3n)^3=27n^3$. Since $\frac{999}{27}=37,$ the cubes we need are $27$ times a cube less than or equal to $37,$ of which there are three. However, $27\cdot1^3=27,$ only has two digits, leaving $\boxed{2}$ such three-digit cubes.
2
Level 2
Number Theory
Problem: How many three-digit perfect cubes are divisible by $9?$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine how many three-digit perfect cubes are divisible by 9, we first identify the range of three-digit numbers. The smallest three-digit number is 100, and the largest is 999. 

Next, we find the smallest integer \( n \) such that \( n^3 \) is at least 100. We calculate \( n = \lceil 100^{1/3} \rceil \). The cube root of 100 is approximately 4.64, so \( n \) must be at least 5. Thus, \( 5^3 = 125 \) is the smallest three-digit perfect cube.

Now, we find the largest integer \( n \) such that \( n^3 \) is at most 999. We calculate \( n = \lfloor 999^{1/3} \rfloor \). The cube root of 999 is approximately 9.997, so \( n \) can be at most 9. Therefore, \( 9^3 = 729 \) is the largest three-digit perfect cube.

The integers \( n \) that yield three-digit perfect cubes are \( n = 5, 6, 7, 8, 9 \). We calculate the cubes of these integers: 
- \( 5^3 = 125 \)
- \( 6^3 = 216 \)
- \( 7^3 = 343 \)
- \( 8^3 = 512 \)
- \( 9^3 = 729 \)

Next, we check which of these cubes are divisible by 9. A number is divisible by 9 if it is a perfect cube of a number that is divisible by 3. Thus, we need to find \( n \) values that are multiples of 3 within our range.

The multiples of 3 between 5 and 9 are 6 and 9. We calculate their cubes:
- \( 6^3 = 216 \) (divisible by 9)
- \( 9^3 = 729 \) (divisible by 9)

The cubes of 5 and 7 are not divisible by 9. Therefore, the only three-digit perfect cubes that are divisible by 9 are 216 and 729.

In conclusion, there are exactly 2 three-digit perfect cubes that are divisible by 9.<|END_OF_LATENT|> The final answer is $2$.