## Task 27/77

Determine the largest natural number $z$ that consists of 10 different digits (in the decimal positional system) and is divisible by 11.
Let

$$
z=10^{9} a+10^{8} b+10^{7} c+10^{6} d+10^{5} e+10^{4} f+10^{3} g+10^{2} h+10 i+k
$$

with $a ; b ; \ldots ; k \in N, a ; b ; \ldots ; k \leq 9, a \neq b \neq \ldots \neq k$ be the number we are looking for. Then, based on the given conditions, the equations

$$
\begin{array}{r}
a+b+c+d+e+f+g+h+i+k=45 \\
a-b+c-d+e-f+g-h+i-k=11 n
\end{array}
$$

$(n \in \mathbb{Z})$ hold. By adding and subtracting these two equations and then dividing by 2, we obtain

$$
\begin{aligned}
a+c+e+g+i & =22+5 n+\frac{n+1}{2} \\
b+d+f+h+k & =22-5 n-\frac{n-1}{2}
\end{aligned}
$$

Due to the integer nature of the sums on the left sides of the equations, $n$ must be odd: $n=2 m+1$, $m \in \mathbb{Z}$. Given

$$
10 \leq a+c+e+g+i \leq 35 \quad ; \quad 10 \leq b+d+f+h+k \leq 35
$$

we have $m=0$. This leads to

$$
a+c+e+g+i=28 \quad ; \quad b+d+f+h+k=17
$$

We now choose, to satisfy the maximality condition, the digits with higher place values to be as large as possible, and obtain step by step

$$
\begin{array}{cc|cc}
a=9 & c+e+g+h=19 & b=8 & d+f+h+k=9 \\
\hline c=7 & e+g+i=12 & d=6 & f+h+k=3 \\
\hline e=5 & g+i=7 & f=2 & h+k=1 \\
\hline g=4 & i=3 & h=1 & k=0
\end{array}
$$

Thus, we get $z=9876524130$. The verification confirms the divisibility by 11.