## Task 18/90

Determine all four-digit numbers (in the decimal system) with the following property: If you multiply them by the number that contains exactly the same digits in reverse order, the result is an eight-digit number divisible by 1000.
The following considerations lead to the solution:

1. Since the product is to be an eight-digit number, the sought numbers cannot end in zero.
2. Since the product is to be divisible by 1000, one factor must be divisible by 125, but not by 250 (due to 1.), and the other must be divisible by 8.
3. Therefore, half of the sought numbers have the form

$$
1000 a+125 \text { or } 1000 a+375 \text { or } 1000 a+625 \text { or } 1000 a+875
$$

with $a \in N, 0<a \leq 9$. 4. For the second factor, this results in

$$
5210+a \text { or } \quad 5730+a \text { or } \quad 5260+a \text { or } \quad 5780+a
$$

5. To achieve divisibility by 8 for the second factors, in the first two cases $a=6$, and in the last two cases $a=4$.
6. Thus, the following 8 numbers satisfy the given conditions:

$6125 ; 6375 ; 4625 ; 4875 ; 5216 ; 5736 ; 5264 ; 5784$. By the solution method, other numbers are excluded.