2016 AMC 10A Problem 5

Write the side lengths as $k$, $3k$, $4k$ for some positive integer $k$. The volume is
$$k \cdot 3k \cdot 4k = 12k^3.$$

So the volume must be $12$ times a perfect cube: $12 \cdot 1 = 12$, $12 \cdot 8 = 96$, $12 \cdot 27 = 324$, and so on. Among the choices, only $96 = 12 \cdot 2^3$ has this form (a $2 \times 6 \times 8$ box).

The answer is $\boxed{\textbf{(D)}\ 96}$.

A ratio with integer parts is parametrized by one scaling integer $k$; every derived quantity becomes a monomial in $k$, and the volume being cubic in $k$ is what filters the choices. Checking divisibility by $12$ alone is not enough; the cofactor must itself be a cube.

• Checking only whether the volume is divisible by 12 (48 and 144 also are) instead of whether the quotient is a perfect cube.
• Allowing non-integer $k$ (for instance $48 = 12 \cdot 4$ with $k = \sqrt[3]{4}$), which violates the integer-side condition.