We built a large, solid cube from more than ten unit cubes, then painted all the faces of the large cube. After that, we separated from the rest those small cubes that have a painted face. Can the number of painted cubes be a multiple of the number of unpainted cubes?
Solution. To get a larger solid cube from more than 10 small cubes, we need to use at least 27 small cubes. This way, we get a $3 \times 3 \times 3$ large cube. Let's paint its faces and start removing the painted faces.

First, remove the "front" and "back" $3 \times 3$ layers. In the remaining $3 \times 3$ layer, only 1 unpainted small cube remains. Since $1 \mid 26$, it is true that the number of painted cubes is a multiple of the number of unpainted cubes.

![](https://cdn.mathpix.com/cropped/2024_05_02_f0d0d52c5f3eed9fe30dg-1.jpg?height=320&width=553&top_left_y=396&top_left_x=775)

In general, if we construct a large cube from $(n+2)^{3}$ small cubes and paint its faces, the number of unpainted cubes will be $n^{3}$. The question is, when will $n^{3}$ be a divisor of $(n+2)^{3}-n^{3}$? If this is true, then $n^{3}$ is also a divisor of $(n+2)^{3}$, which implies that $n$ is a divisor of $(n+2)^{3}$. That is, $n \mid n^{3}+6 n^{2}+12 n+8$. This can only be true if $n \mid 8$, meaning $n=1,2,4$ or 8.

For $n=1$, then $n+2=3$ and $3^{3}-1^{3}=27-1=26$, which is indeed a multiple of 1.

For $n=2, n+2=4, (n+2)^{3}-n^{3}=64-8=56$; here, $8 \mid 56$ is also true.

If $n=4$, then $n+2=6, (n+2)^{3}-n^{3}=216-64=152$, and $64 \nmid 152$. In this case, we do not get a solution.

Finally, if $n=8$, then $(n+2)^{3}-n^{3}=1000-512=488$ and $512 \nmid 488$.

In total, there are two cubes that meet the requirements.

Note. 1. To answer the question of the problem, it was sufficient to show that there exists a painted cube that meets the requirements.

2. From the fundamental theorem of arithmetic (the uniqueness of the prime factorization), it follows that $n^{3}$ is exactly a divisor of $(n+2)^{3}$ if and only if $n \mid n+2$, that is, $n \mid 2$. This shows that the value of $n$ can only be 1 or 2.