Problem 8.5. A grid rectangle of area $S$ is such that:

- it can be completely cut along the grid lines into rectangles $1 \times 13$;
- it can be completely cut along the grid lines into three-cell corners (examples of corners are shown in the figure below);
- there does not exist a grid rectangle of smaller area that satisfies the two previous conditions.

(a) (2 points) Find $S$.

(b) (2 points) What can the perimeter of this rectangle be? List all possible options.
# Answer:

(a) (2 points) 78.

(b) ( points) 38, 58, 82.

Solution. (a) From the condition, it follows that $S$ is divisible by 39. We will prove that $S \neq 39$. Clearly, the rectangle $1 \times 39$ does not work.

Consider the rectangle $3 \times 13$ (3 rows, 13 columns). Suppose it can be cut into three-cell corners. Then the cells of the first column belong to exactly two corners. At the same time, the cells of the second column are also completely covered by the same two corners.

Cut off the first two columns (since they are already cut into two corners). Repeat the same reasoning with the next pair of columns, then the next, and so on. In the end, one last column will remain, which cannot be cut into corners. Contradiction.

For $S=78$, it is not difficult to construct an example - see the next item.

(b) There are several ways to represent $S=78$ as the product of two natural numbers:

- $1 \times 78$
- $2 \times 39$;
- $3 \times 26$
- $6 \times 13$.

The first case is obviously not suitable. In the remaining three cases, we will construct examples of partitions into corners (the partition into rectangles $1 \times 13$ is obvious). The answer to the problem is the perimeters of the three corresponding rectangles.

Case $2 \times 39$ (2 rows, 39 columns). Cut the entire structure into 13 horizontal rectangles $2 \times 3$. Then cut each of them into two corners.

![](https://cdn.mathpix.com/cropped/2024_05_06_1281ceb10571c66f6e2bg-05.jpg?height=51&width=709&top_left_y=1868&top_left_x=372)

Case $3 \times 26$ (3 rows, 26 columns). Cut the entire structure into 13 vertical rectangles $2 \times 3$. Then cut each of them into two corners.

![](https://cdn.mathpix.com/cropped/2024_05_06_1281ceb10571c66f6e2bg-06.jpg?height=71&width=483&top_left_y=211&top_left_x=485)

Case $6 \times 13$ (6 rows, 13 columns). Cut off the rectangle formed by the first three columns. Then cut it into 3 horizontal rectangles $2 \times 3$. The remaining rectangle $6 \times 10$ can easily be cut into 10 vertical rectangles $2 \times 3$. Then cut each of them into two corners.

![](https://cdn.mathpix.com/cropped/2024_05_06_1281ceb10571c66f6e2bg-06.jpg?height=125&width=248&top_left_y=517&top_left_x=599)