In the sense of the task:

$$
\begin{gathered}
100 x+10 y+z=8^{2} \cdot x+8 z+y \\
100 x+10 y+z-16=6^{2} \cdot y+6 x+z \\
100 x+10 y+z+18=4^{2} \cdot z+4 y+x
\end{gathered}
$$

From these equations, \( x=1, y=3, z=9 \); the sought number: 139.

Note. If we swap the digits in the hundreds and tens places, the resulting number is 319. This number cannot actually represent a number in the base-6 system, as the digits in this system are: \(0,1,2,3,4,5\); the number 9 in this system is written as 13. The other conditions in the problem are also not entirely appropriate.

(Géza Sasvári, Pécs.)

The problem was also solved by: Bayer B., Benedek Zs., Bertrám J., Burján K., Faith F., Filkorn J., Freibauer E., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Messik G., Messik V., Perl Gy., Sasvári J., Singer A., Stromfeld F., Szabó J., Szeberényi J.
We are looking for a three-digit integer in the decimal number system, which has the following properties: if the digits in the tens and units place are swapped, the sought number is obtained when written in octal (base-8); if the digits in the hundreds and tens place are swapped, a number is obtained which, when read in base-6, is 16 less than the sought number; finally, by swapping the digits in the hundreds and units place, a number in base-4 is obtained which is 18 greater than the sought number.