## Task 30/74

All pairs of three-digit natural numbers $(x ; y)$ are sought that have the following properties:

1. The numbers of a pair differ in the last two digits exactly by the order of the digits.
2. The difference $x-y$ is 18.
3. Each of the two numbers of a pair is the product of exactly two two-digit prime numbers.
If $x=100 x_{1}+10 x-2+x_{3}$ and $y=100 y_{1}+10 y_{2}+y_{3}$ are the two numbers of one of the sought pairs, then according to condition 1: $y_{2}=x_{3}$ and $y_{3}=x_{2}$.

According to condition 2,

$$
x-y=100 x_{1}+10 x_{2}+x_{3}-\left(100 y_{1}+10 y_{2}+y_{3}\right)=100\left(x_{1}-y_{1}\right)+9\left(x_{2}-x_{3}\right)=18
$$

If both sides of the last equation are divided by 9, we get

$$
11\left(x_{1}-y_{1}\right)+\frac{1}{9}\left(x_{1}-y_{1}\right)+x_{2}-x_{3}=2
$$

This equation is only solvable with single-digit values for $x_{1}, x_{2}, x_{3}, y_{1}$ if $x_{1}=y_{1}$. Then it follows that $x_{2}=x_{3}+2$.

According to condition 3, $x_{2}$ and $x_{3}$ must be odd and not equal to 5. Therefore, only $x_{3}=1 ; x_{2}=3$ and $x_{3}=7 ; x_{2}=9$ are possible. Additionally, according to condition 3, the sum of the digits $x_{1}+x_{2}+x_{3} \neq 0$ (mod 3); since $x_{2}+x_{1} \equiv 1(\bmod 3)$, it follows that $x_{1} \neq 2(\bmod 3)$. Thus, $x_{1}$ can take the values $1 ; 3 ; 4 ; 6 ; 7 ; 9$. Therefore, the following 12 number pairs are initially considered:

$$
\begin{aligned}
& (131 ; 113),(197 ; 179),(331 ; 313),(397 ; 379),(431 ; 413),(497 ; 479) \\
& (631 ; 613),(697 ; 679),(731 ; 713),(797 ; 779),(931 ; 913),(997 ; 979)
\end{aligned}
$$

From these, one first eliminates all number pairs where at least one number is a prime number, as they contradict condition 3. Only the pairs

$$
(697 ; 679),(731 ; 713),(931 ; 913)
$$

remain. If the numbers of these pairs are factored into their prime factors, it is found that the pair (731; 713) is the only pair that satisfies condition 3.