## Task 1 - 030931

We are looking for all three-digit numbers consisting of different digits, where the sum of all two-digit numbers that can be formed from any two of its digits is equal to twice the number.
The sought number $z$ can be written from its digits as follows:

$$
z=100 a+10 b+c \quad(\text { with } 0 \leq a, b, c \leq 9 \text { and } a \neq 0)
$$

Each two-digit number formed from the digits of $z$ is given by $10 x+y$ with $x \in a, b, c, y \in a, b, c$ and $x \neq y$. The equation now reads, with the 6 two-digit numbers formed from the digits of $z$:

$$
2 \cdot z=2 \cdot(100 a+10 b+c)=(10 a+b)+(10 a+c)+(10 b+a)+(10 b+c)+(10 c+a)+(10 c+b)
$$

Summarized, this yields: $200 a+20 b+2 c=22 a+22 b+22 c$ and further: $178 a=2 b+20 c$ or, after simplification: $89 a=b+10 c$.

Estimating the right-hand term from above, i.e., setting $b=9$ and $c=9$, we obtain the inequality $89 a \leq 99$ and thus $a \leq 1$. This implies $a=1$ and yields for the two remaining digits $b=9$ and $c=8$.

The sought number is therefore $z=198$ and is the only number that satisfies the given conditions.