## Task 6 - 210836

Determine all six-digit natural numbers $z$ with the following property:

If the first digit of $z$ is placed at the last position, while the sequence of the remaining five digits remains unchanged, the resulting number $z^{\prime}$ is three times as large as the original number $z$.
I. If a six-digit natural number $z$ has the required property, it follows: If $x$ is the first digit of $z$, then $z=100000 x+y$ with a natural number $y$ for which $y<100000$ holds, and $z^{\prime}=10 y+x$, and it holds that

$$
(100000 x+y) \cdot 3=10 y+x, \quad \text { thus } \quad 42857 x=y
$$

If $x \geq 3$, then $y \geq 42857 \cdot 3$ would contradict $y<100000$.

Therefore (and because $x$ as the first digit of a multi-digit number is greater than 0), only either $x=1, y=42857$ or $x=2, y=42857 \cdot 2=85714$ can be true.

Thus, at most the numbers $z=142857$ and $z=285714$ can have the required property.

They do have this property; because $3 \cdot 142857=428571$ and $3 \cdot 285714=857142$.

Therefore, exactly these two six-digit natural numbers have the required property.

Solutions of the third round 1981 adopted from [5]

### 5.24 XXII. Olympiad 1982

### 5.24.1 First Round 1982, Class 8