Find the digits $A,B,C$ such that $ABC+BAC+CAB=ABBC$

Question

A,B,C are distinct digits of a three digit number such that

                         A B C
                         B A C
                      +  C A B
                     ____________
                        A B B C

Find the value of A+B+C.

a) 16 b) 17 c) 18 d) 19

I tried it out by using the digits 16 17 18 19 by breaking them in three numbers but due to so large number of ways of breaking I cannot help my cause.

Answer 1

The largest possible carry from a sum of three digits is $2$, so from the righthand column we see that $2C+B$ is $C$, $10+C$, or $20+C$, and therefore $C+B$ is $0$, $10$, or $20$. The sum of two distinct digits cannot be $0$ or $20$, so $C+B=10$, and there is a carry of $1$ to the middle column.

From the middle column we then see that $B+2A+1$ is $B$, $10+B$, or $20+B$, so that $2A+1$ is $0$, $10$, or $20$. But all of these are impossible since $2A+1$ is clearly odd. Thus, the problem has no solution.

Answer 2

Let, $A=1, B=2, C=3$

The digit $A$ has $3$ times $=1\cdot 3=3$

The digit $B$ has $3$ times$=2\cdot 3=6$

The digit $C$ has $3$ times $=3\cdot 3=9$
Now:

$A+B+C=3+6+9=18$