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 $$A+B+C\equiv S(A)+S(B)+S(C)=1+2+3+4+5+6+7+8+9+0=45\equiv 0(\mathrm{mod}9).$$ By condition, we have $A+B=C$, so $2C\equiv 0(\mathrm{mod}9)$, whence it follows that $C\equiv 0(\mathrm{mod}9)$, and, therefore, $S(C)\equiv 0(\mathrm{mod}9)$, i.e. the sum of the digits of $C$ is divisible by $9$.

Note that $S(C)=S(A+B)\le S(A)+S(B)$. 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, $S(C)\le S(A)+S(B)$ 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 $18$.