# I. solution: 

1. $b=1$, because the first partial product $c d \cdot b=c d$.
2. $f=0$, because in the second subtraction $c-f=c$.
3. $c \neq 0$ according to the condition $d \neq 0$, otherwise the second partial product $(d f) 0$ would be.
4. From the second partial product, it follows that the value of $c \cdot d$ is a number ending in 0, and adding the tens digit of $c \cdot c$ to it results in a single-digit number $(d)$.

Therefore, (since $c=0, c=1$ are excluded, and for $c>2$ the value of $c^{2}+1$ is already two-digit) only $c=2$ is possible.

5. Since, as we have seen, $c d=2 d 0$ ends in 0, $d$ can only be 5.
6. Therefore, the divisor is 25, the quotient is 125, and thus the dividend is $125 \cdot 25=3125$.

Performing the complete division:

$3125: 25=125$

25

we find that $a=3$, and $e=6$.

Béla Tímár (Pannonhalma, Bencés g. II. o. t.)

II. solution: $\overline{a b c d}=\overline{b c d} \cdot \overline{c d}$, which means $1000 a=\overline{b c d}(\overline{c d}-1) . \overline{b c d}$ and $\overline{c d}-1$ can only be divisible by 5, and only one can be even. This means that one of the factors is divisible by 125 and the other by 8. Only $\overline{a b c}$ can be divisible by 125, so $\overline{b c} 25,50$ or 75 is possible (since $b=c=0$ is excluded). $\overline{c d}=50$ is not possible, because then $\overline{b c d}$ would not be divisible by 8 and $\overline{c d}-1$ would be odd. $\overline{c d}=75$ also does not fit, because 74 is not divisible by 8. Therefore, $\overline{c d}=25, \overline{b c d}=125$ or 625. Since $25 \cdot 625$ is already a five-digit number, $\overline{b c d}=125, \overline{a b c d}=25 \cdot 125=3125$, and the other digits can be determined by performing the division.
Sure, here is the translated text:

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

\[
\begin{aligned}
& a b c d: c d=b c d \\
& \underline{c d} \\
& e c \\
& \frac{d f}{b c} d \\
& \underline{b c d}
\end{aligned}
\]
```