## Task $5 / 80$

How many natural numbers are there whose representation in the decimal system consists exactly of the digits 1, 2, 3, 4, 5, 6, 7, 8 (where each also appears only once) and which are divisible by 11 without a remainder?
As is well known, the rule applies: A natural number $n$ is divisible by 11 without remainder if and only if its alternating digit sum $Q_{A}(n)$ is divisible by 11 without remainder.

If the sum of the four terms in $Q_{A}(n)$ that are preceded by a plus sign is denoted by $s_{1}$ and the sum of the remaining four terms by $s_{2}$, then the following two equations hold:

$$
s_{1}+s_{2}=Q(n)=36 \quad(1) \quad ; \quad s_{1}-s_{2}=Q_{A}(n)=11 k
$$

$(k \in \mathbb{Z})$. The smallest value that the sums $s_{1}$ and $s_{2}$ can take is $1+2+3+4=10$, and the largest is $5+6+7+8=26$. Therefore, $10 \leq s_{1} \leq 26 \quad$ (3).

Adding or subtracting (1) and (2) yields

$$
2 s_{1}=36+11 k \quad ; \quad 2 s_{2}=36-11 k
$$

from which it immediately follows that $k$ must be even. From (3) and (4), it further follows that

$$
20 \leq s_{1}=36+11 k \leq 52 \quad ; \quad-16 \leq 11 k \leq 16
$$

Since $k$ is even, this inequality is only satisfied for $k=0$. The two equations (4) then yield $s_{1}=18, s_{2}=18$. There are now the following possibilities for forming the sums $s_{1}$ and $s_{2}$:

$$
\begin{array}{cccc}
s_{1} & s_{2} & s_{1} & s_{2} \\
1+2+7+8 & 3+4+5+6 & 1+3+6+8 & 2+4+5+7 \\
1+4+5+8 & 2+3+6+7 & 1+4+6+7 & 2+3+5+8 \\
2+3+6+7 & 1+4+5+8 & 2+4+5+7 & 1+3+6+8 \\
2+3+5+8 & 1+4+6+7 & 3+4+5+6 & 1+2+7+8
\end{array}
$$

Since the order in each of the two sums can be chosen in 4! different ways, a total of $8 \cdot 4!\cdot 4!=4608$ different numbers with the required property result.