Belarusian Mathematical Olympiad

a)
For example, $765 + 324 = 1089$.

b)
Since any integer $Y$ is congruent modulo $9$ to the sum of its digits, we have
$$A + B + C \equiv S(A) + S(B) + S(C) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45 \equiv 0 \pmod{9}.$$
By condition, we have $A + B = C$, so $2C \equiv 0 \pmod{9}$, whence it follows that $C \equiv 0 \pmod{9}$, and, therefore, $S(C) \equiv 0 \pmod{9}$, i.e. the sum of the digits of $C$ is divisible by $9$.

Note that $S(C) = S(A+B) \leq S(A) + S(B)$. Indeed, if the sum of the digits of $A$ and $B$ does not exceed $9$ in all number positions, then when we add $A$ and $B$ there is no 'carry' from any number position, so $S(A+B) = S(A) + S(B)$. Otherwise, since the greatest possible carry is $1$, if there is a carry from some number position, then the sum $S(A+B)$ decreases by $9$.

Thus, $S(C) \leq S(A) + S(B)$ and $18$ are smaller than $22.5$ and are divisible by $9$. The following examples show that $S(C)$ can be $9$ (the digits are $1+0+8+9$); $765 + 324 = 1089$ (the sum of the digits is equal to $18$).

Therefore, the possible values for the sum of the digits of $C$ are $9$ or $18$.