7.5. What is the minimum number of 3-cell corners that need to be painted in a $6 \times 6$ square so that no more corners can be painted? (Painted corners must not overlap.)
Answer: 6.

Solution. Let the cells of a $6 \times 6$ square be painted in such a way that no more corners can be painted. Then, in each $2 \times 2$ square, at least 2 cells are painted, otherwise, a corner in this square can still be painted. By dividing the $6 \times 6$ square into 9 $2 \times 2$ squares, we get that at least $9 \cdot 2 = 18$ cells are painted. Therefore, at least 6 corners are painted.

Figure 3 shows how to paint 6 corners so that no more corners can be painted.

Comment. It is proven that the number of painted corners is no less than $6-4$ points.

An example with 6 painted corners is drawn - 3 points.

![](https://cdn.mathpix.com/cropped/2024_05_06_085f4ad889ab6c44b23dg-3.jpg?height=261&width=265&top_left_y=418&top_left_x=1027)

Figure 3