## Task 9 - V00709

Solve the following number puzzle, where the same letters represent the same digits:

$$
\begin{aligned}
& \mathrm{abb}-\mathrm{cdb}=\mathrm{ebd} \\
& \begin{array}{ll}
: & + \\
\mathrm{fg} * & \mathrm{ch}=\mathrm{gic}
\end{array} \\
& \mathrm{fk}+\mathrm{cfc}=\mathrm{cbi}
\end{aligned}
$$
From the top subtraction problem, it immediately follows that $d=1$. Assuming $f=1$, then $a$ can only be 2 or 3. Both are not possible, as $c$ and $e$ must be smaller than $a$ and $f=1$ would be. Assuming $f=3$, then $a=9$ and $g$ and $k$ would be 1 and 2, respectively, or vice versa. $f g=31$ and $f g=32$ are both not possible, as then $c h>45$ and their product would be four-digit. Therefore, $f=2$.

$$
\begin{gathered}
\mathrm{abb}-\mathrm{cOb}=\mathrm{eb0} \\
: \quad+\quad- \\
2 \mathrm{~g} * \mathrm{ch}=\mathrm{gic} \\
----- \\
2 \mathrm{k}+\mathrm{c} 2 \mathrm{c}=\mathrm{cbi}
\end{gathered}
$$

$g$ and $k$ cannot be 1, as then $b$ would be identical to $g$ or $k$. Therefore, $a$ is 5, 6, 7, or 8. Among the products of the factors 23 to 29, only $23 * 28=644$ ends in two identical digits, which are 4. Thus, $a=6$ and $b=4$. $g$ and $k$ are 3 and 8, respectively.

```
644-c04 = e40
    :
    2g * ch = gic
    2k + c2c = c4i
```

Assuming $g=8$, a contradiction arises because $g<e<a=6$. Therefore, $g$ is 3 and $k=8$. For $c$, only 1 remains, as otherwise the middle multiplication problem would not be solvable.

```
\(644-104=\) e 40
    :
    \(23 * 1 \mathrm{~h}=3 \mathrm{i} 1\)
    \(28+121=14 i\)
```

The remaining digits are then $e=5, h=7$, and $i=9$.

```
\(644-104=540\)
    :
    \(23 * 17=391\)
    \(28+121=149\)
```

The problem is uniquely solvable.