2. Does a ten-digit number exist that is divisible by 11 and has all different digits on each of its positions?

翻译完成，保留了原文的换行和格式。
Solve Write a three-digit number that can be divided by 11 and has different digits on each place, such as 275, 396, 418. Using these three numbers, it is not difficult to construct a ten-digit number that can be divided by 11. For example:
$$
\begin{array}{l}
2753964180 \\
=275 \cdot 10^{7}+396 \cdot 10^{4}+418 \cdot 10 \\
=11 \cdot\left(25 \cdot 10^{7}+36 \cdot 10^{4}+38 \cdot 10\right) .
\end{array}
$$

Another solution Utilize the divisibility rule for 11, the number
$$
n=\overline{a_{1} a_{2} a_{3} \cdots a_{3}} \text { and } S(n)=a_{1}-a_{2}+a_{3}
$$
$-\cdots-a_{n}$ are simultaneously divisible or not divisible by 11.
Let $A$ be the sum of the numbers in $S(n)$ with a “+” sign, and $B$ be the sum of the numbers with a “-” sign. Clearly,
$A+B=1+2+\cdots+9+0=45$. At the same time, $A-B$ should be divisible by 11. We can take $A-B=11$. Then $A=28, B=17$. It is not difficult to select five different digits that sum to 17, such as $1+2+3+5+6=17$. Then $4+7+8+9=28$. Therefore, the number 1427385960 meets the conditions of the problem.