## Task 6 - 030616

Given are nine squares with side lengths $a=36 \mathrm{~mm}, d=20 \mathrm{~mm}, g=14 \mathrm{~mm}, b=30 \mathrm{~mm}, e=18 \mathrm{~mm}, h=8 \mathrm{~mm}, c=28 \mathrm{~mm}, f=16 \mathrm{~mm}$, $i=2 \mathrm{~mm}$.

Arrange these squares to form a rectangle! To do this, make a drawing!
If one adds the areas of the 9 squares, the rectangle has an area of

$$
A=\left(2^{2}+8^{2}+14^{2}+16^{2}+18^{2}+20^{2}+28^{2}+30^{2}+36^{2}\right) \mathrm{mm}^{2}=4224 \mathrm{~mm}^{2}
$$

4224 has the prime factors 2, 3, and 11, since $4224=2^{7} \cdot 3 \cdot 11$ holds. Assuming natural numbers as the lengths of the rectangle sides and considering that each side must be at least $36 \mathrm{~mm}$ long (otherwise the first square would not fit), the sides could be $44 \mathrm{~mm}$ and $96 \mathrm{~mm}$, or $48 \mathrm{~mm}$ and $88 \mathrm{~mm}$, or $64 \mathrm{~mm}$ and $66 \mathrm{~mm}$ long.

A side length of $44 \mathrm{~mm}$ is not possible, as one can obtain a rectangle side with the $36 \mathrm{~mm}$ square and the $8 \mathrm{~mm}$ square, but the resulting gap at the $8 \mathrm{~mm}$ square cannot be closed. A similar consideration rules out the side length of $48 \mathrm{~mm}$ (with the $30 \mathrm{~mm}$ square and the $18 \mathrm{~mm}$ square).

The sought rectangle has side lengths of $64 \mathrm{~mm}$ and $66 \mathrm{~mm}$.

If one places the two largest squares ($36 \mathrm{~mm}$ and $30 \mathrm{~mm}$) along one side, a solution can be found with some trial and error (see figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0163.jpg?height=417&width=403&top_left_y=1887&top_left_x=1363)