6.4. Workers were laying a floor of size $n \times n$ using two types of tiles: $2 \times 2$ and $2 \times 1$. It turned out that they managed to completely cover the floor such that the same number of tiles of each type was used. For which $n$ could this have been possible? (Cutting tiles or overlapping them is not allowed.)
Answer. For $n$ divisible by 6.

Solution. Let the workers use $x$ tiles of each type. Then the area covered by the tiles is $4x + 2x = 6x = n^2$. Therefore, $n^2$ must be divisible by 2 and 3. Consequently, $n$ must be divisible by 2 and 3, and thus by 6.

If $n$ is divisible by 6, then the floor can be laid. It is sufficient to note that a $2 \times 3$ rectangle can be laid out using two tiles - one of each type. And a square $6k \times 6k$ can be divided into $2 \times 3$ rectangles.

Comment. It is shown that $n$ is divisible by $6$ - 3 points.

It is shown that for $n = 6k$, the floor can be laid - 4 points.

It is shown that for $n = 6$, the floor can be laid - 1 point.

The statement "If $n^2$ is divisible by 6, then $n$ is divisible by 6" can be used without proof.