2007 AMC 10A Problem 17

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 $\boxed{\textbf{(D)}\ 60}$.

"Make this a perfect cube" is a statement about exponents in the prime factorization: pad each exponent up to the next multiple of $3$. The minimal padding gives the minimal $m$, and since $m$ grows cubically while $n$ grows linearly, the minimal $m$ also minimizes $m + n$.

• Stopping at $n = 15$ or $m = 45$ and answering that number, or trying larger $n$ without realizing $m$ grows like a cube.
• Padding to a perfect square instead of a cube (taking $m = 3$, which makes $75m = 225 = 15^2$, not a cube).