6. From a paper square $8 \times 8$, p seven-cell corners were cut out. It turned out that it was impossible to cut out any more such corners. For what smallest $n$ is this possible? A seven-cell corner is obtained by cutting out a $3 \times 3$ square (in cells) from a $4 \times 4$ square.
Answer: $\mathrm{n}=3$.
Instructions. Example. The figures show two examples of placing three corners so that no more can be cut out.
![](https://cdn.mathpix.com/cropped/2024_05_06_0c37f0795934e72a769eg-2.jpg?height=622&width=1248&top_left_y=338&top_left_x=1522)

Cells belonging to the same corner are marked with the same number. Other examples are possible.

Estimate. It remains to show that if two corners have been cut out, then another one can be cut out. If there is a row and a column with no cells of the corners, then taking the cell at their intersection as the "vertex" of the corner and directing the sides of the corner towards the more distant sides, we get a corner that can be cut out. If such a row and column do not exist, then let the columns be covered (otherwise, rotate the board by $90^{\circ}$). In this case, there are two rectangles $1 \times 4$ covering the columns. Cutting the board along the horizontal axis of symmetry, we get a corner in each part. In one of the parts (in any), there will be a rectangle adjacent to the short edge, with no cells of the corner, and at least four cells away from the covering rectangle. One of its edge cells serves as the vertex of the corner that can be cut out.

Criteria. Example without estimate: 3 points.