4. consider three side squares of an $n \times n \times n$ cube that meet at one of the corners of the cube. For which $n$ is it possible to cover them completely and without overlapping with paper strips of size $3 \times 1$? The paper strips can also be glued over the edges of the cube.

## Solution
If $n$ is divisible by 3. Insert coordinates in each side face so that $(1,1)$ is the field in the common corner and $(n, n)$ is the diametral corner field. Color all fields $(a, b)$ with $a, b \not \equiv 2(\bmod 3)$ black and the rest white. Then each strip of paper covers an even number of black squares and so the total number of black squares must be even. However, a quick calculation shows that the number of black fields per side area for $n=3 m+1$ and $n=3 m+2$ is equal to $(2 m+1)^{2}$, i.e. odd. This means that the total number of black fields is also odd, which is a contradiction. So $n$ must be divisible by 3. In this case, however, each side can be covered individually with paper strips.