1. Among all ten-digit numbers divisible by eleven, in which no digit is repeated, find the smallest and the largest.
SOLUTION. Let the considered ten-digit number be denoted by $\overline{a_{9} a_{8} a_{7} a_{6} a_{5} a_{4} a_{3} a_{2} a_{1} a_{0}}$, where $a_{9}$, $a_{8}, \ldots, a_{0}$ are distinct digits, i.e., all digits $0,1,2, \ldots, 9$ in some order. Further, let $s_{2}=a_{0}+a_{2}+a_{4}+a_{6}+a_{8}$ be the sum of its digits at even ${ }^{1}$ positions $^{2}$ and $s_{1}=a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ be the sum of digits at odd positions.

To determine divisibility by eleven, we use the known criterion: The number $\overline{a_{9} a_{8} \ldots a_{1} a_{0}}$ is divisible by eleven if and only if the corresponding difference $s_{2}-s_{1}$ is divisible by eleven. Clearly, $\left|s_{2}-s_{1}\right| \leqq(9+8+7+6+5)-(4+3+2+1+0)=25$ or $-25 \leqq s_{2}-s_{1} \leqq 25$. The sum $s_{2}+s_{1}=0+1+2+\ldots+9=45$ is an odd number, so the number $s_{2}-s_{1}$ must also be odd. Therefore, for a suitable number, there are two possibilities: $s_{2}-s_{1}=-11$ or $s_{2}-s_{1}=11$.

In the first case, from the system of equations $s_{2}+s_{1}=45, s_{2}-s_{1}=11$ we get $s_{2}=28$, $s_{1}=17$, and in the second case, $s_{2}=17$, $s_{1}=28$.

The number 17 can be expressed in all possible ways as the sum of five distinct digits:

$$
\begin{aligned}
17 & =9+5+2+1+0=9+4+3+1+0= \\
& =8+6+2+1+0=8+5+3+1+0=8+4+3+2+0= \\
& =7+6+3+1+0=7+5+4+1+0=7+5+3+2+0=7+4+3+2+1= \\
& =6+5+4+2+0=6+5+3+2+1 .
\end{aligned}
$$

Among the ten-digit numbers written with all ten digits, the largest are certainly those that start with the digits 987 or even 9876. Given the found decompositions of the number 17, this is clearly not possible for $s_{1}=17$, but it is for $s_{2}=17$: it is enough to take the sum $17=8+6+2+1+0$ for $s_{2}$, which is also the only possibility. The remaining digits will then be uniquely completed based on this choice to get the largest possible number. The sought largest number is therefore 9876524130.

The smallest number can be found by an analogous procedure. Since $a_{9} \neq 0$, the smallest among the considered numbers are certainly those that start with the digits 102. From the found decompositions of the number 17, we see again that this can be achieved only by the choice $s_{1}=17=6+5+3+2+1$. This corresponds to the number (since we know all its digits at odd and even positions, their arrangement is determined by the requirement that the resulting number be the smallest) 1024375869.

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