2005 AMC 10A Problem 11

Count the red faces and the total faces separately, then compare.

1. Red faces. Paint only touches the outside of the big cube. Each of its $6$ faces is an $n$ by $n$ grid of little squares, so there are $6 \cdot n^2$ red faces. Cutting does not add or remove any paint.
2. All faces. There are $n^3$ small cubes, each with $6$ faces, so there are $6 \cdot n^3$ faces in all.
3. The fraction that is red.
$$\frac{6n^2}{6n^3} = \frac{1}{n}.$$
4. Match it to one fourth. $\frac{1}{n} = \frac{1}{4}$, so $n = 4$.

Check: a $4 \times 4 \times 4$ cube gives $64$ small cubes with $64 \times 6 = 384$ faces. The red surface has $6 \times 16 = 96$ squares, and $96 \div 384 = \frac{1}{4}$.

The answer is $\boxed{\textbf{(B)}\ 4}$.

The red faces are exactly the painted surface of the original cube, cut into small squares. Surface grows like $n^2$ while the number of small cubes grows like $n^3$, so the red share is always $\frac{1}{n}$. Counting the painted area as a whole is much quicker than sorting cubes by how many red faces each one has.