Let's determine the digits $A, B, C, D$ such that $A B C \cdot A D=A D D C$.

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On the left side, at least $1000 A^{2}$ appears, while on the right side, at most $1000(A+1)-1$, so $A^{2}<A+1$. Therefore, $A=1$ is the only possibility, since $A$ is a positive integer.

Substituting this into the equation

$$
(100 A+10 B+C)(10 A+D)=1000 A+100 D+10 D+C
$$

after performing the operations, we arrive at the equality

$$
100 D+100 B+10 B D+10 C+C D=110 D+C
$$

From this,

$$
100 B+10 B D+10 C+C D=10 D+C
$$

follows. The number on the right side of (2) is less than 100, so only $B=0$ is possible. Substituting this, we get the equality

$$
10 C+C D=10 D+C
$$

Rewriting it: $10(D-C)=C(D-1)$. The number on the left side is divisible by 5, which is only possible if one of the factors on the right side is a multiple of 5. Since neither $C$ nor $D-1$ can be greater than 9, four cases are possible: $D-1=0, D-1=5, C=5$ or $C=0$.

In the first case, the multiplication is $101 \cdot 11=1111$, in the second case $104 \cdot 16=1664$, in the third case $105 \cdot 19=1995$, and finally $100 \cdot 10=1000$.

Thus, these four multiplications satisfy the requirements of the problem.

Remark. Some of the submitters assumed from the outset - in accordance with the customs - that different letters represent different numbers. Thus, they only got 2 solutions. These solutions were worth 4 points. Lesson: Always read the problem statement carefully, and do not imagine anything that is not written there.