## Task 24/84

Determine all four-digit natural numbers n (in the decimal system) that satisfy the following conditions:

1. The cross sum is 23.
2. The alternating cross sum is -5.
3. The cross product is 360.
4. The sum of the digits in the first and third positions is equal to the digit in the second position.
Assuming there is (at least) one number

$$
n=10^{3} a+10^{2} b+10 c+d
$$

which satisfies the given conditions $(a ; b ; c ; d \in N, a ; b ; c ; d \leq 9)$. Then the following equations hold:

$$
\begin{array}{r}
a+b+c+d=23 \\
a-b+c-d=-5 \\
a b c d=360 \\
a-b+c=0
\end{array}
$$

By adding (1) and (2), and using (4), we get: $a+c=9=b$ (5), and by substituting into (1), $d=5$. From (3), $a c=8$ (6).

Since $a ; c \in N$, it follows immediately from (5) and (6) that $a_{1}=1, c_{1}=8$ and $a_{2}=8, c_{2}=1$. The verification confirms that the two numbers $n_{1}=1985$ and $n_{2}=8915$ satisfy all the given conditions. No further numbers $n$ can exist based on the solution method.

### 2.25 Problems and Solutions 1985