1. Is it possible to place 0 or 1 in each cell of a $5 \times 5$ square table so that the sum in each $2 \times 2$ square is divisible by 3, while both zeros and ones are present in the table?
Answer: Yes, it is possible.

Solution. For example, as follows:

| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |

Note. This arrangement is not unique.

Criteria. A correct arrangement without an explanation of how it is obtained - 7 points.