2009 AMC 10A Problem 11

1. Try one cube first to see what happens. Side $4$: the cube has volume $64$. The new box is $5 \times 3 \times 4 = 60$. The drop is $4$, which is the side length.
2. Now in general. Call the side $s$, so the cube has volume $s^3$. The new box measures $s + 1$, $s - 1$ and $s$.
3. First multiply $(s + 1)(s - 1) = s^2 - s + s - 1 = s^2 - 1$.
4. So the new volume is
$$s(s^2 - 1) = s^3 - s.$$
It is always exactly $s$ less than the cube.
5. The problem says the drop is $5$, so $s = 5$, and the cube's volume is $5^3 = 125$.
6. Check: $6 \times 4 \times 5 = 120$, and $125 - 120 = 5$.

The answer is $\boxed{\textbf{(D)}\ 125}$.

Making one side $1$ longer and another $1$ shorter multiplies them to $s^2 - 1$ instead of $s^2$. That "minus $1$" times the third side $s$ is the whole loss. Spotting $(s+1)(s-1) = s^2 - 1$ turns a cube problem into a one-step equation.

Every choice is a cube: $2^3, 3^3, 4^3, 5^3, 6^3$. Test the sides. Side $5$ gives a box $6 \times 4 \times 5 = 120$, which is $5$ less than $125$. The other sides give drops of $2, 3, 4, 6$, never $5$.

Let the cube have side $s$, so its volume is $s^3$. The new solid has dimensions $s + 1$, $s - 1$ and $s$, so its volume is
$$s(s+1)(s-1) = s(s^2 - 1) = s^3 - s.$$

This is exactly $s$ less than the cube's volume. The problem says the difference is $5$, so $s = 5$ and the cube's volume is
$$5^3 = 125.$$

Check: the new box is $6 \times 4 \times 5 = 120 = 125 - 5$.

The answer is $\boxed{\textbf{(D)}\ 125}$.