Problem 217

Perfect cubes have the property that the number of times that each prime factor occurs is a multiple of 3. This is because its prime factors can be separated into three identical groups; in this case, the product of each group is the cube root of the original number.

In particular, if $n^3 = 2^4 \times 3^2 \times 5^5 \times k$ where $n$ is an integer, then the number of times that the prime factors 2, 3 and 5 occur in the integer $n^3$ must be multiples of 3.

Since $n^3$ already includes 4 factors of 2, then $k$ must include at least 2 additional factors of 2, so that $n^3$ has a factor of $2^6$. ($k$ could also include more factors of 2, as long as the total number of factors of 2 is a multiple of 3.)

Since $n^3$ already includes 2 factors of 3, then $k$ must include at least 1 additional factor of 3.

Since $n^3$ already includes 5 factors of 5, then $k$ must include at least 1 additional factor of 5.

Therefore, $k$ includes at least 2 factors of 2, at least 1 factor of 3, and at least 1 factor of 5.

This means that the smallest possible value of $k$ is $2^2 \times 3 \times 5 = 60$. In principle, $k$ could also include other prime factors, but to make $k$ as small as possible, we do not need to consider this further.