Question

In the following addition, A, B, C represent different digits. Find them and the sum

Answer 1

Here we observe that the last digit of $A+B+C$ is C, so that $A+B=10$. Since C is a digit, $C\le 9$. Hence the carry from unit's place to ten's place is $1$.
Since we are adding only $3$ digits, the carry in ten's place of the sum cannot exceed $2$. Hence $B$ cannot be more than $2$. Thus we observe that $B=1$ or $2$. The addition of digits in ten's place gives (along with carry $1$ from unit's place) $A+B+C+1=10+C+1$ and this must leave remainder $A$ when divided by $10$.
If $B=1$, then $A=9$ and hence $C+1=9$ giving $C=8$. We get $99+11+88=198$, which is a correct answer. If $B=2$, you get $A=8$ and $C+1=8$ giving $C=7$. But then $88+22+77=187$. This does not fit in as hundred's place in the sum is $1$ but not $2$.
The correct answer is $99+11+88=198$.