A2. Marko divided a large rectangle into 7 smaller rectangles, the lengths of whose sides $\mathrm{v}$ meters are natural numbers. $\mathrm{Na} 5$ of these rectangles, he wrote down their area (see the figure). What is the minimum number of square meters of the combined area of the remaining 2 rectangles?
(A) 36
(B) 54
(C) 64
(D) 76
(E) 81
A2.

![](https://cdn.mathpix.com/cropped/2024_06_07_e347356273e0c1397da7g-18.jpg?height=402&width=371&top_left_y=887&top_left_x=251)

Let's denote the lengths of the sides of the two smaller remaining rectangles in meters as $a, b, c$, and $d$, as shown in the figure. We observe that the width of the large rectangle is at most $24 \, \text{m}$. Since $21 + 14 > 24$, it must be that $b > 1$. At the same time, $b$ must divide 21 and 14, as the sides of the rectangles are natural numbers in meters. Since the greatest common divisor of 21 and 14 is 7, it follows that $b = 7$. Similarly, we conclude that $d > 1$ and $d$ divides 10 and 35, so $d = 5$. The width of the large rectangle in meters is thus $a + \frac{21}{b} + \frac{14}{b} = a + 5$ and at the same time $\frac{10}{d} + c + \frac{35}{d} = c + 9 > 9$. The combined area of the two remaining rectangles will be the smallest when $a$ and $c$ are the smallest possible, which means the width of the large rectangle will be the smallest possible. Since this width is greater than 9 and must divide 24, it is at least 12. In this case, $a = 7$ and $c = 3$, and the combined area of the two remaining rectangles is $a b + c d = 7 \cdot 7 + 3 \cdot 5 = 64 \, \text{m}^2$.