Let's determine the four-digit number $A B C D$ such that the following division is correct: $A B C D: D=D B A$. (Different letters represent different digits.)





Translating the text as requested, while preserving the line breaks and format.
Let's write the condition in product form:

$$
\overline{D B A} \cdot D = \overline{A B C D}
$$

First, consider the digits $A$ and $D$; as leading digits, neither can be 0.

The last digit of the product, which is the last digit of the partial product $D \cdot A$, is equal to $D$. Therefore, the difference $D A - D = D(A-1)$ is divisible by 10. Hence, at least one of $D$ and $A-1$ must be divisible by 5, since 5 is a prime number. If $D$ is divisible by 5, then $D$ can only be 5. If $A-1$ is divisible by 5, then its value can be 5 or 0; thus, the following possibilities arise:

$$
\text{a) } D = 5, \quad \text{b) } A = 6, \quad \text{c) } A = 1
$$

On the other hand, for the multiplicand:

$$
100 D + 10 B + A \geq 100 D + 10 \cdot 0 + 1 = 100 D + 1, \quad \text{so} \quad 100 D + 1 \geq 100 \cdot 1 + 1 = 101
$$

If $D = 5$, then $100 D + 1 = 501$, which is greater than 101. Therefore, $A$ must be 2, but then $D(A-1) = 5$, which is not divisible by 10. This assumption does not lead to a solution.

In the case of b) (6), from the condition $D^2 \cdot 10 A - D = 60 - D \geq 52, \quad D > 7$, only $D = 8$ is possible. However, from the requirement:

$$
\begin{gathered}
(800 + 10 B + 6) \cdot 8 = 6000 + 100 B + 10 C + 8 \\
2 B + C = B + B + C = 44
\end{gathered}
$$

This is impossible because the sum of three digits is at most 27. Thus, we do not get a solution from this case either.

In the remaining case $A = 1$, from (6) we have $D^2 \cdot 10 A - D \geq 10 - 4 = 6, \quad D \geq 3$.

The possible values for $D$ are 3 and 4. From (2):

$$
\begin{gathered}
100 D^2 + 10 B D + D = 1000 + 100 B + 10 C + D \\
C + (10 - D) B = 10(D^2 - 10)
\end{gathered}
$$

The left side is never negative, but the right side is only non-negative if $D > 3$.

For the remaining possibility $D = 4$:

$$
C + 6 B = 60
$$

Here, $C \leq 9$, so $6 B \geq 51$, and $B > 8$. With $B = 9$ and $C = 6$, we get a solution:

$$
491 \cdot 4 = 1964
$$

Thus, the sought number is 1964.

In the solution, we did not need to use the fact that different letters must be replaced by different digits, only that the leading digit cannot be 0.