In the following division (where identical letters represent identical digits), replace the letters with digits so that the equality is correct:

| abcd : ef $=$ fga |
| :--- |
| hb |
| dic |
| be |
| dhd |
| dhd |
From $b-b=i$ it follows that ..... $\mathbf{i}=\mathbf{0}$
From $d i-b=d$ it follows that ..... $\mathbf{d}=\mathbf{1}$
$d i-b=d$, that is, $10-b=1$, and thus $b$ can only be 9 or 8. But $b$ is the last digit of $f^{2}$ and a square number cannot end in 8, so $\ldots$ ..... $\mathbf{b}=\mathbf{9}$
$f^{2}$ ends in 9, so $f=3$ or 7. The product of ef and $f$ is a two-digit number, so $f$ cannot be 7, because even $17 \cdot 7=119$ is a three-digit number, so ..... $\mathbf{f}=\mathbf{3}$
$f e=h$, that is, $3 e=h$. But 1,3 are already used, so ..... $\mathrm{e}=\mathbf{2}$
and ..... $\mathbf{h}=\mathbf{6}$
From $a-h=d$, that is, $a-6=1$ ..... $\mathrm{a}=\mathbf{7}$
From $c-e=h$, that is, $c-2=6$ ..... $\mathrm{c}=8$
From $e f \cdot g=b e$, that is, $23 \cdot g=92$ ..... $\mathbf{g}=\mathbf{4}$
With the knowledge of the letters, we can write:

$$
\begin{aligned}
& 7981: 23=347 \\
& \frac{69}{108} \\
& 92
\end{aligned}
$$