4. A $10 \times 10$ square was cut into 17 rectangles, each with both side lengths greater than 1. What is the smallest number of squares that could be among these rectangles? Provide an example of such a cutting.
Answer: one square.

Solution. Suppose that among the rectangles, there is not a single square. Let $a$ and $b$ be the sides of an arbitrary rectangle, with $a > b$. Since the integer $b$ is greater than 1, then $b \geqslant 2$, and thus $a \geqslant 3$. Therefore, the area of each such rectangle $a \times b$ is at least $2 \cdot 3 = 6$ cells. However, then 17 rectangles should occupy at least $17 \cdot 6 = 102$ cells, while the original square $10 \times 10$ has only 100 cells.

In Figure 1, an example of cutting the $10 \times 10$ square into 17 rectangles, among which there is exactly one square, is provided.

Criteria. Only the answer - 0 points. Proved that the area of each non-square rectangle is at least 6 cells - 1 point. Proved,

![](https://cdn.mathpix.com/cropped/2024_05_06_cbe5f382c3da7be3217bg-2.jpg?height=452&width=456&top_left_y=365&top_left_x=1515)

Fig. 1 that among the rectangles there is a square - another 3 points. Correct example of cutting - another 3 points. Full solution - 7 points.