2009 AMC 10A Problem 11

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 $\boxed{\textbf{(D)}\ 125}$.

Stretching one dimension by $1$ and shrinking another by $1$ changes the volume by the difference-of-squares factor $(s+1)(s-1) = s^2 - 1$, and the "$-1$" times the untouched side $s$ is the whole loss. Recognizing $(s+1)(s-1)$ instantly turns a cubic-looking condition into a linear one.

The choices are the cubes $2^3, 3^3, 4^3, 5^3, 6^3$. Test each side: $s = 5$ gives $6 \cdot 4 \cdot 5 = 120$, five less than $125$. The other sides give losses of $2, 3, 4, 6$, never $5$.

• Expanding (s+1)(s-1) as s^2 - 2s - 1 or s^2 + 1, or forgetting to cube the side length and answering 5.
• Changing the same dimension twice (using $s \cdot s \cdot s$ or $(s+1)(s-1)(s-1)$) instead of three different dimensions.