5. Misha has a $7 \times 7$ square of paper, all cells of which are white. Misha wants to color $N$ cells black. What is the smallest $N$ for which Misha can color the cells so that after coloring, no completely white rectangle with at least ten cells can be cut out from the square?
Answer: 4.

Solution. Divide the $7 \times 7$ square into 5 rectangles: four $3 \times 4$ rectangles (each corner of such a rectangle coincides with one of the corners of the $7 \times 7$ square) and a $1 \times 1$ square. If only three cells are colored, there will be a white rectangle consisting of 12 cells. Example for 4 cells: color $b 4, d 2, d 6, f 4$ in chess notation.

Criteria. Only the answer - 0 points. Correct example for 4 cells - 3 points. Proof that 3 black cells are not enough - 4 points. Points for parts of this problem are summed.