Problem 7.1.1. Let $A, B, C, D, E, F, G, H$ - be different digits from 0 to 7 - satisfy the equation

$$
\overline{A B C}+\overline{D E}=\overline{F G H}
$$

Find $\overline{D E}$, if $\overline{A B C}=146$.

(The notation $\overline{A B C}$ represents a three-digit number consisting of digits $A, B, C$, similarly constructed are the numbers $\overline{F G H}$ and $\overline{D E$).
Answer: 57.

Solution. Substitute 146 for $\overline{A B C}$ and write the example in a column:

$$
\begin{array}{r}
146 \\
+\quad D E \\
\hline F G H
\end{array}
$$

$D, E, F, G, H$ can only be the digits $0,2,3,5,7$. Let's consider $E$.

- If $E=0$, then $H=6$ - but this contradicts the fact that $C=6$.
- If $E=2$, then $H=8$ - but we don't have such a digit.
- If $E=3$, then $H=9$ - but we don't have such a digit.
- If $E=5$, then $H=1$ - but this contradicts the fact that $A=1$.

By the method of elimination, we understand that $E=7$ :

$$
\begin{array}{r}
146 \\
+\quad D 7 \\
\hline F G 3
\end{array}
$$

Since numbers cannot start with zero, it can only correspond to one digit - this is the digit $G$ :

$$
\begin{array}{r}
146 \\
+\quad D 7 \\
\hline F 03
\end{array}
$$

Now it is not difficult to fully restore the example:

| 146 |
| ---: |
| $+\quad 57$ |
| 203 |