The rectangular grid in the figure is made of 31 segments of $0.5 \mathrm{~cm}$ and comprises 12 squares. Rosa drew on a rectangular sheet of $21 \mathrm{~cm}$ by $29.7 \mathrm{~cm}$, grid-lined with squares of side $0.5 \mathrm{~cm}$, a large rectangular grid made with 1997 segments. How many squares does this rectangle have?

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-06.jpg?height=228&width=301&top_left_y=1339&top_left_x=1432)
Let $m$ and $n$ be, respectively, the number of segments of $0.5 \, \text{cm}$ on two consecutive sides of the rectangle. We know that the total number of segments of $0.5 \, \text{cm}$ in the division of the rectangle into $m \times n$ squares of side $0.5 \, \text{cm}$ is: $m(n+1) + n(m+1)$ (prove this). Thus,

$$
m(n+1) + n(m+1) = 1997 \Rightarrow n = \frac{1997 - m}{2m + 1}
$$

Moreover, one of the sides considered is less than or equal to the other, say: $m \leq n$. In this case, we can conclude that $m \leq 31$, because

$$
n \geq m \Rightarrow n(m+1) + m(n+1) \geq 2m(m+1)
$$

Thus, $1997 \geq 2m(m+1)$ and since $1998 > 1997$, it follows that

$$
1998 > 2m(m+1) \Rightarrow 999 > m(m+1)
$$

From this, we conclude that $m < 32$.

On the other hand, we have

$$
n = \frac{1997 - m}{2m + 1} \Rightarrow 2n = \frac{3994 - 2m}{2m + 1} = \frac{3995 - (2m + 1)}{2m + 1} \Rightarrow 2n = \frac{3995}{2m + 1} - 1
$$

Thus, the problem now reduces to investigating the divisors of $3995 = 5 \times 17 \times 47$. The only values of $m$ that satisfy the condition $1 \leq m \leq 31$ are $m = 2$, $m = 8$, and $m = 23$, which correspond, respectively, to the divisors 5, 17, and 47. For these values of $m$, we have $n = 399$, $n = 117$, and $n = 42$ respectively. The other divisors will give equivalent configurations (by swapping $m$ and $n$).

Therefore, Rosa could have constructed 3 different configurations with the 1997 segments. The first with $2 \times 399$ squares, the second with $8 \times 117$ squares, and the third with $23 \times 42$ squares.