5. A $7 \times 7$ board has a chessboard coloring. In one move, you can choose any $m \times n$ rectangle of cells and repaint all its cells to the opposite color (black cells become white, white cells become black). What is the minimum number of moves required to make the board monochromatic?

Answer: in 6 moves.
Solution. Consider segments of length equal to the side of a cell, separating pairs of cells adjacent to the side of the board. Along each side of the board, there are 6 such segments, totaling $6 \cdot 4=24$. Each of these segments separates cells that initially have different colors, so each segment must end up on the side of at least one of the chosen rectangles. It is easy to see that no more than four of the considered segments can lie on the sides of one rectangle. Thus, to make the board monochromatic, at least $24: 4=6$ moves are required. It remains to show that six moves are sufficient. We choose 6 rectangles of size $1 \times 6$, covering the cells of the second, fourth, and sixth rows and the second, fourth, and sixth diagonals.

Criteria for evaluation.

Only the answer - 0 points.

An example of repainting in 6 moves without proving minimality - 2 points.

Proof of minimality without an example - 5 points.