2. In the ternary number, the hundreds are greater than the tens, and the tens are greater than the units. If the digits of this ternary number are written in reverse order and the resulting number is added to the original, the resulting number contains only odd digits. Determine all ternary numbers for which this is true.
II/2. Let's denote the three-digit number as $\overline{a b c}$. It must hold that $a > b > c$, and the number $\overline{a b c} + \overline{c b a} = 10^2(a + c) + 10(2b) + (a + c)$ is composed of only odd digits. If $a + c < 10$, the tens digit is even, which is not possible. Therefore, $a + c \geq 10$. Since $a + c$ is also odd, it follows that $a + c \geq 11$. We can write $a + c = 10 + l$, where $l$ is an odd number. Then,

$$
\overline{a b c} + \overline{c b a} = 10^3 + 10^2 l + 10(2b + 1) + l
$$

From this, it follows that $2b + 1 < 10$, because otherwise the hundreds digit would be even. Therefore, $2b < 9$ and thus $b \leq 4$.

Since $c < b \leq 4$, it follows that $c \leq 3$. Therefore, $a + c \leq 9 + 3 = 12$. We have already shown that $a + c$ is odd and at least 11, so $a + c = 11$. If $c = 2$ and $a = 9$, then $b$ can be 3 or 4. If $c = 3$ and $a = 8$, then $b$ can only be 4.

The valid numbers are 843, 932, and 942.
Observation that $a$ and $c$ have different parities ..... 1 point
Observation that $a + c \geq 10$ ..... 1 point
Observation that $b \leq 4$ ..... 1 point
Observation that $c \leq 3$ ..... 1 point
Each individual solution ..... $1 + 1 + 1$ points