7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (a total of $65 \times 65$ cells, which is 4225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess?
# Solution

Cells should be colored in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be colored. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table larger than 4x4 such that the knight visits each cell exactly once (see, for example, a 5x5 table). If these moves are numbered, it is clear that fewer than $\left[\frac{N^{2}}{2}\right]$ cells cannot be colored because then there would necessarily be two consecutive uncolored cells in this sequence, i.e., a move from one to the other would be possible. The $35 \times 35$ table needs to be divided into 169 $5 \times 5$ tables, and they should be numbered sequentially, starting with the first $5 \times 5$ table. $\left[\frac{65^{2}}{2}\right]=2112$.

| 21 | 16 | 5 | 10 | 23 |  |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 6 | 11 | 22 | 17 | 4 |  |
| 1 | 20 | 15 | 24 | 9 | 26 |
| 12 | 7 | 18 | 3 | 14 |  |
| 19 | 2 | 13 | 8 | 25 |  |
|  |  |  |  |  |  |

Answer: 2112.