## Task 35/69

We are looking for a natural number $n$ with a four-digit decimal representation, which has the following properties:

1. Its cross sum is an odd square number.
2. It is the product of exactly two different prime numbers.
3. The sum of the two prime numbers is ten times the number obtained by removing the units digit and the tens digit from the number $n$.
4. The difference between one prime factor and ten times the other is equal to the number obtained by removing the hundreds digit and the thousands digit from the number $n$.
Since the cross sum of a four-digit number can be at most 36, only the cross sum 25 is possible for n; because the cross sums 1 and 9 contradict property 2. Let $n = 1000a + 100b + 10c + d$, where $a, b, c, d$ are natural numbers with $1 \leq a \leq 9, 0 \leq b; c; d \leq 9$. From the given properties, we further obtain

$$
\begin{aligned}
n = p_{1} \cdot p_{2} & \text{ (Property 2)} \\
p_{1} + p_{2} = 10(10a + b) & \text{ (Property 3)} \\
p_{2} - 10p_{1} = 10c + d & \text{ (Property 4)} \\
a + b + c + d = 25 & \text{ (Property 5)}
\end{aligned}
$$

Thus, we get

$$
n = p_{1} p_{2} = 1000a + 100b + 10c + d = 1 p(p_{1} + p_{2}) + (p_{2} - 10p_{1}) = 11p_{2} \rightarrow p_{1} = 11
$$

The number sought is thus divisible by 11 without a remainder; this is the case if and only if the alternating cross sum is divisible by 11. This must therefore be -11, 0, or +11. This results in the systems

$$
\begin{array}{r}
a + b + c + d = 25; -a + b - c + d = -11 \\
a + b + c + d = 25; -a + b - c + d = 0 \\
a + b + c + d = 25; -a + b - c + d = +11
\end{array}
$$

System (2) leads to a contradiction by adding the two equations, that an even number is equal to an odd number, and can thus be ruled out as unsolvable.

From System (1), we get $a + c = 18; b + d = 7$, from which $a = c = 9$ and the following values for $b$ and $d$ result: $b = 0; 1; 2; 3; 4; 5; 6; 7, d = 7; 6; 5; 4; 3; 2; 1; 0$. None of the resulting numbers $n$ have the required properties.

From System (3), we finally get $a + c = 7, b + d = 18$ with $b = d = 9$ and $a = 1; 2; 3; 4; 5; 6; 7$ as well as $c = 6; 5; 4; 3; 2; 1; 0$. Of the numbers $n$ formed from these, only $n = 1969$ has all the required properties.