## Task B-4.6.

Determine the three-digit number whose representation in base 11 consists of the same digits as its representation in base 9, but in reverse order.
## Solution.

We are looking for a number for which $x y z_{(9)}=z y x_{(11)}$, where $x, y, z \in\{0,1,2,3,4,5,6,7,8\}$ (digits in base 9), $x \neq 0, z \neq 0$.

By expanding, we get

$$
\begin{gathered}
x \cdot 9^{2}+y \cdot 9+z=z \cdot 11^{2}+y \cdot 11+x \\
81 x+9 y+z=121 z+11 y+x \\
80 x-2 y-120 z=0 \\
y=40 x-60 z=20(2 x-3 z)
\end{gathered}
$$

Since $y$ is a digit, the only possibility is $y=0$.

Thus $2 x-3 z=0$, i.e., $2 x=3 z$.

The number $z$ must be even.

Let's examine all possibilities:

from $z=2$ follows $x=3: 302_{(9)}=203_{(11)}(=245)$ 1 point

from $z=4$ follows $x=6: 604_{(9)}=406_{(11)}(=490)$ 1 point

for $z \geqslant 6$ it would be $x \geqslant 9$ so there are no other solutions.