11.5. A $7 \times 7$ checkered board was assembled using three types of figures (see the image), not necessarily all. How many figures, composed of four cells, could have been used?

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-4.jpg?height=123&width=463&top_left_y=310&top_left_x=1476)
Answer: only one.

Solution. We will prove that only one figure consisting of four cells can be used. We will color the cells of the board as shown in Fig. 11.5a: Each of the given figures can cover no more than one shaded cell, therefore, the number of figures must be no less than 16.

Since 16 three-cell figures cover 48 cells, one such figure consisting of four cells must be used. More than one cannot be used.

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-4.jpg?height=368&width=371&top_left_y=501&top_left_x=1528)
Indeed, if at least two figures consisting of four cells are used, and the remaining 14 figures are three-cell ones, together they will occupy \(2 \cdot 4 + 14 \cdot 3 = 50\) cells, which exceeds the size of the board.

An example of a board composed of one four-cell figure and fifteen three-cell figures is shown in Fig. 11.56.

There are other examples as well. They can be obtained by highlighting 7 rectangles of size \(3 \times 2\) on the board, each of which can be divided into two three-cell corners, and placing one three-cell and one four-cell figure in the remaining part of the board.

Evaluation criteria.

«+» A complete and justified solution is provided

![](https://cdn.mathpix.com/cropped/2024_05_06_51c28cf2652dc1e7d646g-4.jpg?height=388&width=428&top_left_y=1005&top_left_x=1491)

Fig. 11.56 «士» It is proven that there are no fewer than 16 figures and the correct answer is given, but the example is missing

«Ғ» Only the correct answer and the correct example are provided

«-» Only the answer is provided

«-» The problem is not solved or is solved incorrectly