The integer $m$ is a perfect cube exactly when it is equal to $n^{3}$ for some integer $n$. For example, 1000 is a perfect cube since $1000=10^{3}$. What is the smallest positive integer $k$ for which the integer $2^{4} \times 3^{2} \times 5^{5} \times k$ is a perfect cube?
(A) 12
(B) 30
(C) 60
(D) 480
(E) 1620
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.

ANSWER: (C)