## Task A-1.4.

From nine identical rectangles whose length and width are natural numbers, a rectangular plate of dimensions $20 \times 9$ has been composed. What are all the possible dimensions of the original rectangles?
## Solution.

Since we are using 9 congruent rectangles, and the total area of the board is $9 \cdot 20$, we conclude that each rectangle has an area of 20.

Since the rectangles have integer dimensions, we can discard all dimensions of the rectangles except for these 3 possibilities: $1 \times 20$, $2 \times 10$, and $4 \times 5$.

![](https://cdn.mathpix.com/cropped/2024_05_30_0bfcef02cf5258b8afa5g-03.jpg?height=366&width=814&top_left_y=862&top_left_x=558)

The dimensions $4 \times 5$ are possible because 9 such rectangles can be arranged such that 4 are placed side by side along the length of the board, and the remaining rectangles are placed perpendicular to them.

![](https://cdn.mathpix.com/cropped/2024_05_30_0bfcef02cf5258b8afa5g-03.jpg?height=377&width=811&top_left_y=1436&top_left_x=545)

The dimensions $2 \times 10$ are not possible because such rectangles can only tile a board whose width and length are even numbers.

Note: For the dimensions $1 \times 20$ and $4 \times 5$, the student must provide an example. It is sufficient to draw or describe the example in words (both are not required).

Note: Students can also show that the dimensions $2 \times 10$ are not possible in the following way.

Since the length of the board is 9, rectangles with dimensions $2 \times 10$ can only be placed such that their side of length 10 is parallel to the side of the board of length 20. (1 point)

Therefore, we would have to place two such rectangles end to end to form a rectangle $2 \times 20$. After arranging 4 such pairs of rectangles, a remaining part of the board with dimensions $1 \times 20$ would be left, which cannot be tiled with the remaining one rectangle. (1 point)