## Problem 2

\begin{aligned} AAA \\
BBB \\
AAAC \\
\end{aligned}

all A’s denote somoe digit, all B’s denote another idigt and C denotes a third digit. What are these digits?

A and B are being added in all columns but the result differs in the last column from the first and second column. That means the sum have a carry that’s being applied to the first and second column.

Adding any two numbers from 1 to 9, can only end up giving a carry of 1 (e.g. 8+2, 7+3, 6+4). Knowing that, we can guess that the first A of the result is a 1, therefore all A’s are ones

\begin{aligned} 111 \\
BBB \\
111C \\
\end{aligned}

Now, B must be 9, as only 9+1 give us a carry. And this makes C = 0.

\begin{aligned} 111 \\
999 \\
1110 \\
\end{aligned}