2. The representation of a natural number $n$ in bases 7 and 9 has the same three digits, but in reverse order. Determine all such numbers.
## Solution.

The condition of the problem can be written as: $(a b c)_{7}=(c b a)_{9}$, that is,

$$
7^{2} a+7 b+c=9^{2} c+9 b+a
$$

After rearranging, we get $b=8(3 a-5 c)$.

It follows that $8 \mid b$, but since $b<7$ ( $b$ is a digit in base 7), it must be $b=0$, so $3 a=5 c$,

that is (since $a$ and $c \in\{0,1,2,3,4,5,6\}$), $a=5$ and $c=3$ (the solution $a=b=0$ would give zero, but zero is not a natural number).

The desired number is $503_{7}=305_{9}=248_{10}$.

Note. A student who, in addition to the correct solution, also provides solutions for $b=8$ (which yields $a=2, c=1$ and $a=7, c=4$) loses 6 points.