Let $a, b, c$ be natural numbers, all greater than zero and less than or equal to 9. Then the three-digit number and the numbers resulting from rearranging its digits are:

$$
\begin{array}{ll}
100 a+10 b+c & 100 a+10 c+b \\
100 b+10 a+c & 100 b+10 c+a \\
100 c+10 a+b & 100 c+10 b+a
\end{array}
$$

The sum $s$ is then:

$$
\begin{aligned}
s=100(2 a+2 b+2 c)+10(2 a & +2 b+2 c)+(2 a+2 b+2 c)=(100+10+1) \cdot(2 a+2 b+2 c)= \\
& =111 \cdot 2 \cdot(a+b+c)=111 \cdot 2 Q
\end{aligned}
$$

Example: Let the number be 534. Then $2 Q=24$ and $2 Q \cdot 111=24 \cdot 111=2664$ or $534+543+354+345+$ $453+435=2664$.
## Task 2 - 070722

Horst says to Klaus: Name a three-digit natural number, none of whose digits are zero and no two are the same! Write it down and list all the three-digit numbers that can be formed by rearranging the digits of the given number! Add all these numbers!

Before Klaus was finished, Horst had already found the result in his head. He calculated: $2 Q \cdot 111$, where $Q$ is the sum of the digits of the first number mentioned.

Explain his method in general and then give a numerical example!