Problem 10.1. The entire surface of a cube $13 \times 13 \times 13$ was painted red, and then this cube was sawn into smaller cubes $1 \times 1 \times 1$. All the faces of the smaller cubes $1 \times 1 \times 1$ that were not painted red were painted blue. By what factor is the total area of the blue faces greater than the total area of the red faces?
Answer: 12.

Solution. Each face of the original cube consists of exactly $13^{2}$ $1 \times 1$ squares, so a total of $6 \cdot 13^{2}$ $1 \times 1$ squares were painted red. Since there are exactly $13^{3}$ $1 \times 1 \times 1$ cubes, and each has 6 faces, the number of $1 \times 1$ squares painted blue is $6 \cdot 13^{3}-6 \cdot 13^{2}$. Therefore, the answer to the problem is

$$
\frac{6 \cdot 13^{3}-6 \cdot 13^{2}}{6 \cdot 13^{2}}=13-1=12
$$

Remark. The same answer could have been obtained in a different way, by understanding that each $1 \times 1$ red square on the surface of the original cube corresponds to exactly 12 blue $1 \times 1$ squares inside the cube. These 12 blue squares are obtained from the red one by applying a parallel translation of length 1 "into the cube," perpendicular to the face of the red square.