## Task 4 - 130734

Determine all three-digit natural numbers $a$ that are equal to half the sum of the two numbers that arise from cyclic permutation of the digits of $a$!

Hint: If the number $a$ is represented by the digit sequence $u v w$, then the numbers $v w u$ and $w u v$ arise from cyclic permutation. Cases with $v=0$ or $w=0$ should also be allowed; the numbers resulting from cyclic permutation do not have to be three-digit.
Assuming a number with the digit sequence $x y z$ meets the conditions of the problem. Then $a=100 x+10 y+z$ is equal to half the sum of $b=100 y+10 z+x$ and $c=100 z+10 x+y$. Therefore, we have:

$$
200 x+20 y+2 z=2 a=b+c=101 y+110 z+11 x
$$

Thus, $189 x=81 y+108 z$ and hence

$$
7 x=3 y+4 z
$$

Therefore, 7 is a divisor of $3 y+4 z$ and thus also of $3 y+4 z-7 z=3(y-z)$, and hence of $y-z$.

Given $0 \leq y \leq 9,0 \leq z \leq 9$, it follows that either $y=z$ and then $y=z=x \geq 1$ according to (1), or $z$ is 7 greater than $y$.

Therefore, only the possibilities for $y$ and $z$ listed in the following table remain, to which only the specified $x$ belongs in each case due to (1):

| $\mathrm{x}$ | $\mathrm{y}$ | $\mathrm{z}$ | $\mathrm{x}$ | $\mathrm{y}$ | $\mathrm{z}$ | $\mathrm{x}$ | $\mathrm{y}$ | $\mathrm{z}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 2 | 2 | 2 | $\cdots$ | $\cdots$ | $\cdots$ |
| 9 | 9 | 9 | 8 | 1 | 4 | 7 | 0 | 3 |
| 0 | 7 | 4 | 1 | 8 | 5 | 2 | 9 | 6 |
| 9 | 2 | 5 |  |  |  |  |  |  |

Therefore, the numbers 111, 222, 333, 444, 555, 666, 777, 888, 999, 592, 481, 370, 407, 518, 629 can be solutions at most. For 111, ..., 999, this is immediately clear; furthermore, we have

$$
\begin{array}{ll}
\frac{1}{2} \cdot(925+259)=1184: 2=529 & \frac{1}{2} \cdot(814+148)=962: 2=481 \\
\frac{1}{2} \cdot(703+37)=740: 2=370 & \frac{1}{2} \cdot(74+740)=814: 2=407 \\
\frac{1}{2} \cdot(185+851)=1036: 2=518 & \frac{1}{2} \cdot(296+962)=1258: 2=629
\end{array}
$$

Thus, each of these 15 numbers satisfies the conditions of the problem.