2007 AMC 10A Problem 17

A perfect cube, like $8 = 2 \times 2 \times 2$ or $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$, has each prime factor appearing a multiple of $3$ times. So look at the prime factors of $75$.

1. Factor. $75 = 3 \times 5 \times 5$. There is one $3$ and two $5$s.
2. What is missing? To reach three of each, we need two more $3$s and one more $5$. So the smallest $m$ is
$$m = 3 \times 3 \times 5 = 45.$$
3. Check. $75 \times 45 = 3375 = 15 \times 15 \times 15$. So $n = 15$.
4. Sum. $m + n = 45 + 15 = 60$.
5. Could a bigger $n$ give a smaller sum? No. Any $n$ that works must be a multiple of $15$ (it needs a $3$ and a $5$). If $n = 30$, then $m = 30^3 \div 75 = 360$, which is much bigger. Making $n$ bigger makes $m$ grow very fast, so the smallest choice wins.

The answer is $\boxed{\textbf{(D)}\ 60}$.

"Make this a perfect cube" is really about counting prime factors: each prime must show up a multiple of $3$ times. The smallest $m$ just fills in what is missing. Because $m$ grows like a cube as $n$ grows, the smallest $m$ also gives the smallest sum.

Factor: $75 = 3 \cdot 5^2$. For $75m$ to be a perfect cube, every prime in its factorization must appear to a power divisible by $3$. The prime $3$ currently has exponent $1$, so $m$ must contribute at least $3^2$; the prime $5$ has exponent $2$, so $m$ must contribute at least $5^1$.

The smallest choice is $m = 3^2 \cdot 5 = 45$, which gives $75 \cdot 45 = 3^3 \cdot 5^3 = 15^3$, so $n = 15$ and $m + n = 60$.

Could a larger $n$ give a smaller sum? Any valid $n$ must be divisible by $3$ and $5$, so $n = 15k$, and then $m = \frac{(15k)^3}{75} = 45k^3$. The sum $45k^3 + 15k$ increases with $k$, so $k = 1$ is optimal.

The answer is $\boxed{\textbf{(D)}\ 60}$.