## Task 11/75

We are looking for all positive three-digit numbers with the following property: If you multiply the number by the number obtained by reversing the sequence of digits, you get a six-digit product that ends with two zeros.
The sought numbers cannot end with a zero, as otherwise no six-digit number would result as a product. The largest product that can be formed according to the given rule with a number ending in zero is $990 \cdot 099 < 990 \cdot 100 = 99000 < 100000$.

Since the product is divisible by 100, one of the two factors must be divisible by 25, and the other by 4. Therefore, one of the two factors ends with 25 or 75 (since 50 is not possible), and the last two digits of the other factor must be divisible by 4. If we denote the first digit of the first factor with x, we can initially establish the following products:

$$
\text { (1) } \quad x 25 \cdot 52 x \quad ;(2) \quad x 75 \cdot 75 x
$$

It is evident that in case (1) $x=4$ or $x=8$, and in case (2) $x=2$ or $x=6$. This yields the following numbers: (1) $425,524,825,528 ; \quad$ (2) $275,572,675,756$.