Shapovalov A.V.

Each face of a cube $6 \times 6 \times 6$ is divided into $1 \times 1$ cells. The cube is covered with $2 \times 2$ squares such that each square covers exactly four cells, no squares overlap, and each cell is covered by the same number of squares. What is the maximum value that this identical number can take? (Squares can be bent over the edge.)
Evaluation. A cell in the corner of a face can be covered in three ways (entirely within the face, with a fold over one edge of the corner, with a fold over the other edge of the corner). This means that each cell is covered by no more than three squares.

Example. Consider the usual covering of a cube with squares, where each face is covered by nine squares. From the usual covering, we can obtain a rotated one: leave two opposite faces untouched, and on the other four faces, shift all the squares in a ring by one cell. Since a pair of opposite faces can be chosen in three ways, there will be exactly three rotated coverings.

We will show that no two squares on coverings rotated differently coincide. Indeed, consider one face. On the diagram, the centers of the squares covering it are marked: with crosses if this face was not shifted, with black and white dots - if it was shifted in one or the other direction. It is clear that no centers, and therefore no squares, coincide.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3ab7e582e0ed8c4b2bbg-22.jpg?height=449&width=443&top_left_y=1656&top_left_x=802)

## Answer

3.