Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:

a) in 5 or fewer;

b) in 4 or fewer;

![](https://cdn.mathpix.com/cropped/2024_05_06_611ab840a0f62c05d9afg-08.jpg?height=309&width=311&top_left_y=110&top_left_x=1011)

c) in 3 or fewer such foldings?

(If yes, write the number of the fold after which each cell will be shaded for the first time, draw the fold line and mark it with the same number. If no, prove it.) [4 points] (T. I. Golenishcheva-Kutuzova, M. A. Raskin, I. V. Yashchenko)
Answer. It is possible (even with 3 bends).

Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)

Comment. It is impossible to paint all cells in 2 bends. Indeed, with each bend, the number of cells increases by no more than 2 times. Initially, there are 6 painted cells; therefore, after two bends, no more than $6 \cdot 2 \cdot 2=24$ cells out of 36 will be painted.

| 3 | 3 | 3 | 3 | 3 | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 3 | 3 | 3 | 3 | 3 | 3 |
| 3 | 3 | 3 | 3 | 3 | 3 |
|  | 1 |  | 1 | 2 | 2 |
| 2 |  | 1 | 2 | 2 |  |
| 1 | 2 | 2 |  |  | 2 |
|  |  | 1 | 2 |  |  |