Let's find the pairs of four-digit (natural) numbers that are derived from each other by writing the digits in reverse order, and whose product ends in three 0s.
I. solution. Let the two numbers in decimal notation be $\overline{A B C D}$ and $\overline{D C B A}$. Their product is divisible by $1000=2^{3} \cdot 5^{3}$, and thus by 10, but neither of the unit digits $D$ and $A$ is 0, because they are the leading digits in the other number, so one of $D$ and $A$, say $A=5$, and the other, $D$, is one of the digits 2, 4, 6, 8. Thus, $\overline{A B C D}$ is not divisible by 5, so $\overline{D C B A}$ is divisible by $5^{3}=125$, and $\overline{A B C D}$ is divisible by $2^{3}=8$. Both divisibilities can be recognized from the last three digits, so $\overline{C B A}$ is one of the numbers 125, 375, 625, and 875, and $D$ is the digit that, when written at the end of the reverse of these numbers, makes the last three digits of the four-digit number $\overline{521 D}, \overline{573 D}, \overline{526 D}$, or $\overline{578 D}$ divisible by 8. In the first case, only 216 is divisible by 8 between 212 and 218, so $D=6$, and one solution is $5216 \cdot 6125$. Similarly, one suitable pair of numbers results from each of the other three cases: $5736 \cdot 6375, 5264 \cdot 4625$, and $5784 \cdot 4875$. There are no other solutions.

Bulkai Tamás (Győr, Benedek-rendi Czuczor G. g. I. o. t.)

II. solution. Using the above notation, we consider those terms of the expansion of $\left(10^{3} A+10^{2} B+10 C+D\right)\left(10^{3} D+10^{2} C+10 B+A\right)$ in which the exponent of 10 is less than 3:

$$
10^{2}(D C+C B+B A)+10(D B+C A)+D A
$$

and we look for conditions on the digits so that the last term is divisible by 10, the sum of the last two terms is divisible by 100, and the entire expression is divisible by $1000$.

Let $A=5$ again, and $D=2 k$, where $k$ is one of the numbers 1, 2, 3, 4. With these, the sum of the last two terms,

$$
10[k(2 B+1)+5 C]
$$

is only divisible by 100 if $2 B+1$ is divisible by 5, that is, if $B=2$ or 7, and since in this case the expression is $50(k+C)$, or $100 k+50(k+C)$, if the bracket is even, $k+C=2 m$.

We still need to examine under which conditions (1) is divisible by 20:

$(B=2:) \quad 5(C+4)+(4 C+1) k=4(k C+C+5)+k+C, \quad$ and

$(B=7:) \quad 5(3 C+14)+(4 C+3) k=4(k C+3 C+17)+3(k+C)+2$.

The conditions for divisibility by 5 and 4 are: for $B=2$: $4 C+1=5 p, k+C=4 q$ (the latter includes $k+C=2 m$), which is satisfied when $C=1$ and $k=3$, that is, $D=6$, and when $C=6$ and $k=2, D=4$; for $B=7$: $4 C+3=5 p, 3(k+C)+2=4 q$, from which similarly $C=3, k=3, D=6$, and $C=8, k=2, D=4$. Again, we obtained the above 4 solutions.