3. Sasha marked several cells in an $8 \times 13$ table such that in any $2 \times 2$ square, there was an odd number of marked cells. Then he marked a few more cells, as a result of which in each $2 \times 2$ square, there became an even number of marked cells. What is the smallest total number of cells that Sasha could have marked?
Answer: 48.

Solution. See example in the figure (the digit 1 is in the cells marked the first time, the digit 2 - the second time)

Estimate. In the table, 24 independent 2x2 squares can be placed. In the first round, at least one cell in each of them was marked. Since each of them ended up with an odd number of marked cells, and after the second round, the number of marked cells became even, at least one cell in each of them was marked again in the second round. In total, we get a minimum of 48 marked

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

cells.