10. (Kyiv City Mathematics Competition Question) There is a three-digit number written in base seven. If the digits are written in reverse order and considered as a three-digit number in base nine, and these two numbers are equal, find this number.
10. Let the number we are looking for be $A$. In base seven, $A$ can be written as $A=7^{2} \cdot a+7 \cdot b+c$, where $a, b, c \in\{0,1,2,3,4,5,6\}, a \neq 0$.
The number $A$ in base nine can be expressed as $A=9^{2} \cdot c+9 \cdot b+a$.
Thus, from the problem, we get the equation $7^{2} \cdot a+7 \cdot b+c=9^{2} \cdot c+9 \cdot b+a$, which simplifies to $b=8(3 a-5 c)$.
Therefore, $b$ is a multiple of 8, and since $b \in\{0,1,2, \cdots, 6\}$, it must be that $b=0$.
From $3 a-5 c=0$ and $a, c \in\{0,1,2, \cdots, 6\}, a \neq 0$, we get $a=5, c=3$.
Thus, the number $A$ in base seven is $(503)_{7}$, in base nine is $(305)_{9}$, and in base ten is $A=7^{2} \cdot 5+3=248$.