4. Determine all natural numbers of the form $\overline{a b}$ knowing that the sum of the digits of the number $\overline{a b}$ is equal to the sum of the digits of the number $5 \cdot \overline{a b}$.

Daniela and Nicolae Stănică, Brăila
## Solution:

Case I. $b=\operatorname{even} \Rightarrow 5 \cdot \overline{a b}=\overline{x 0}$ or $5 \cdot \overline{a b}=\overline{y z 0}$

We have $4 \cdot \overline{a b}=9 \cdot(x-a) \Rightarrow \overline{a b}: 9$, since $a+b=x$ or

$$
4 \cdot \overline{a b}=9 \cdot(11 y+z-a) \Rightarrow \overline{a b}: 9 \text {, since } a+b=y+z
$$

Case II. $b=$ odd $\Rightarrow 5 \cdot \overline{a b}=\overline{x 5}$ or $5 \cdot \overline{a b}=\overline{y z 5}$

We have $4 \cdot \overline{a b}=9 \cdot(x-a) \Rightarrow \overline{a b}: 9$, since $a+b=x+5$ or

$$
4 \cdot \overline{a b}=9 \cdot(11 y+z-a) \Rightarrow \overline{a b}: 9, \text { since } a+b=y+z+5
$$

Conclusion $\overline{a b} \in\{18,27,36,45,54,63,72,81,90,99\}$