1. The area of a rectangle is $\overline{3 x x 3}$, and the lengths of its adjacent sides are $\overline{x x}$ and $x^{2}$. Determine the digit $x$.
First solution.

The lengths of adjacent sides of the rectangle are $\overline{x x}$ and $x^{2}$, so $x \neq 0$, i.e., $x \in\{1,2, \ldots, 9\}$.

The area of the rectangle is $\overline{x x} \cdot x^{2}=\overline{3 x x 3}$.

1 POINT

The product $\overline{3 x x 3}$ is an odd number, so it is sufficient to consider $x \in\{1,3,5,7,9\}$.

2 POINTS

The last digit of the product $\overline{3 x x 3}$ is 3:

- For $x=1$, the last digit of the product $11 \cdot 1^{2}$ is different from 3, so $x \neq 1$. 1 POINT
- For $x=3$, the last digit of the product $33 \cdot 3^{2}$ is different from 3, so $x \neq 3$. 1 POINT
- For $x=5$, the last digit of the product $55 \cdot 5^{2}$ is different from 3, so $x \neq 5$. 1 POINT
- For $x=9$, the last digit of the product $99 \cdot 9^{2}$ is different from 3, so $x \neq 9$. 1 POINT

By checking $77 \cdot 49=3773 \quad 2$ POINTS

we get the solution $x=7$. 1 POINT

TOTAL 10 POINTS

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