## Task $8 / 80$

We are looking for all natural numbers $n$ with the following properties:

1. It is $n=p_{1} \cdot p_{2}$; the product of two (proper) two-digit prime numbers $p_{1}$ and $p_{2}$.
2. For the cross sum $Q(n)$, it holds that $Q(n)=p_{1}$ with $p_{1}<p_{2}$.
3. The units digits of $p_{1}$ and $p_{2}$ are the same.
4. Also $p_{1} \pm 6$ is a (proper) two-digit prime number.
Given that $p_{1}$ and $p_{2}$ are two-digit prime numbers, $n$ is obviously at most four digits. This implies $Q(n) \leq 4 \cdot 9 = 36$. Since $Q(n) = p_{1}$, it follows that $Q(n) = p_{1} \leq 31$.

Therefore, the only possible values for $p_{1}$ are the seven prime numbers $11, 13, 17, 19, 23, 29,$ and $31$. Due to condition 4, the five prime numbers $11, 13, 19, 29,$ and $31$ can be immediately ruled out, leaving only $p_{11} = 17$ and $p_{12} = 23$.

Because of the equality of the units digits, $p_{21}$ can only be the prime numbers 37, 47, 67, and 97, and $p_{22}$ can only be 43, 53, 73, and 83. By using $Q(n) = Q(p_{1} \cdot p_{2}) = p_{1}$, six pairs can be quickly ruled out through trial. Indeed, the two remaining pairs $p_{11} = 17, p_{21} = 37$ and $p_{12} = 23, p_{22} = 73$ satisfy all the conditions of the problem:

$$
\begin{array}{rlr}
17 \cdot 37 = 629 = n_{1}; \quad Q(n_{1}) = 6 + 2 + 9 = 17; & 17 + 6 = 23; & 17 - 6 = 11 \\
23 \cdot 73 = 1679 = n_{2}; \quad Q(n_{2}) = 1 + 6 + 7 + 9 = 23; & 23 + 6 = 29; & 23 - 6 = 17
\end{array}
$$

Thus, there are exactly two natural numbers $n_{1} = 629$ and $n_{2} = 1679$ with the required properties.