B4. The letters $A, B$ and $C$ stand for different, non-zero digits.
Find all the possible solutions to the word-sum shown.
$$
\begin{array}{r}
A B C \\
B C A \\
+\quad C A B \\
\hline A B B C
\end{array}
$$
Solution
Looking at the units column, we can deduce that, as $A$ and $B$ are non-zero and $A+B<20$, $C+A+B=10+C$. So $A+B=10$ and there is a carry of 1 to the tens column. In this column, as $A$ and $C$ are non-zero and $A+C<19,1+B+C+A=10+B$. Hence $C+A=9$ and there is a carry of 1 to the hundreds column. The calculation in the hundreds column is the same as that in the tens column and there is a carry of 1 to the thousands column.
Therefore $A=1, B=10-1=9$ and $C=9-1=8$.
So there is exactly one possible solution to the word-sum: $198+981+819=1998$.