Problem 9.8. In an $8 \times 12$ table, some $N$ cells are black, and the rest are white. In one operation, it is allowed to paint three cells forming a three-cell corner to white (some of them could already be white before repainting). It turned out that it is impossible to make the entire table white in fewer than 25 such operations. Find the smallest possible value of $N$.
Answer: 27.

Solution. Divide the $8 \times 12$ table into 24 squares of $2 \times 2$ (Fig. 9a).

![](https://cdn.mathpix.com/cropped/2024_05_06_a324d57f9e21fe03fd09g-4.jpg?height=349&width=506&top_left_y=875&top_left_x=171)

(a)

![](https://cdn.mathpix.com/cropped/2024_05_06_a324d57f9e21fe03fd09g-4.jpg?height=345&width=505&top_left_y=877&top_left_x=759)

(b)

Fig. 9: to the solution of problem 9.8

First, we will show how to color 27 cells black in the required manner. In each $2 \times 2$ square, color the top-left cell, and in the bottom-right square, also color all the other cells (Fig. 9b). Note that no three-cell corner can simultaneously cover black cells from two different $2 \times 2$ squares. Therefore, it will take at least 2 operations for the bottom-right square, and another 23 operations for the remaining squares, i.e., a total of at least 25 operations.

Now, we will show that if there are 26 black cells, the table can be guaranteed to be made completely white in no more than 24 operations (if there are initially fewer than 26 black cells, some white cells can be considered black). For this, it is sufficient to either once repaint 3 black cells to white, or twice repaint 2 black cells to white. Indeed, after such actions, the number of available operations will be no less than the remaining number of black cells, so it will be enough to repaint at least one black cell each time.

Since there are a total of 26 black cells, and there are 24 $2 \times 2$ squares, either there is a square with 3 black cells, or there are at least two squares with 2 black cells each (otherwise, there would be no more than $24+1=25$ black cells). By repainting the corresponding three-cell corners, and then repainting at least one black cell each time, we will achieve the required result in no more than 24 operations.