4. Pete is trying to lay out a square on a table using identical cardboard rectangles measuring $14 \times 10$. Will he be able to do this? Propose your version of constructing such a square. What is the minimum number of rectangles he will need?
Solution. Let $n$ be the number of rectangles that form the square. Then the area of the square is $S=14 \cdot 10 \cdot n$. If the side of the square is formed by $m$ sides of length 14 and $k$ sides of length 10, then the length of the side of the square is $14 m + 10 k$, and its area is $S=(14 m + 10 k)^{2}$. Equating the expressions for the areas, we get

$$
14 \cdot 10 \cdot n = (14 m + 10 k)^{2} \rightarrow 7 \cdot 5 \cdot n = (7 m + 5 k)^{2}
$$

We can assume that the numbers $m$ and $k$ are coprime. Otherwise, the side of the square can be reduced by a factor of $p > 1$, where $p$ is a common divisor of $m$ and $k$.

The left-hand side of the equation (*) is divisible by 7, so the number $7 m + 5 k$ is divisible by 7, and $(7 m + 5 k)^{2}$ is divisible by 49. Therefore, the left-hand side of (*) is divisible by 49, which is possible only if $n$ is divisible by 7. Similarly, it can be shown that $n$ is divisible by 5. Thus, $n = 35 \cdot s, s \in \mathbb{Z}$, and the smallest possible number of rectangles is $n = 35$. The figure below shows a square assembled from 35 rectangles.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ced405473fd649332a3g-2.jpg?height=603&width=662&top_left_y=298&top_left_x=731)

Answer: 35 rectangles.