$11 \cdot 25$ has a three-digit number when written in base 7. If the digits are written in reverse order and considered as a three-digit number in base 9, and these two numbers are equal. Find this number.
(Kyiv Mathematical Olympiad, 1956)
[Solution] Let the number we are looking for be $A$.
In base 7, $A$ can be written as
$$
A=7^{2} \cdot a+7 \cdot b+c,
$$

where $a, b, c \in\{0,1,2,3,4,5,6\}, \quad a \neq 0$.
The number $A$ in base 9 can be expressed as
$$
A=9^{2} \cdot c+9 \cdot b+a .
$$

Thus, from the problem statement, we get the equation

which is
$$
\begin{array}{c}
7^{2} \cdot a+7 \cdot b+c=9^{2} \cdot c+9 \cdot b+a, \\
b=8(3 a-5 c) .
\end{array}
$$

Thus, $b$ is a multiple of 8, and since $b \in\{0,1,2, \cdots, 6\}$, it can only be
$$
b=0 \text {. }
$$

From
and
$$
3 a-5 c=0
$$

we get
$$
\begin{array}{l}
a, c \in\{0,1,2, \cdots, 6\}, \quad a \neq 0, \\
a=5, \quad c=3 .
\end{array}
$$

Thus, the number $A$ in base 7 is $(503)_{7}$, in base 9 is $(305)_{9}$, and in base 10 is
$$
A=7^{2} \cdot 5+3=248 \text {. }
$$