Given the set $M=|A|$ where $A$ is a ten-digit positive integer with all distinct digits, and $11111|A|$. Find $|M|$.

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Solution: Since the digits of $A$ are all different, we have
$A \equiv 0+1+\cdots+9 \equiv 0(\bmod 9)$, i.e., $9 \mid A$.
Also, $11111 \mid A$, and since $(9,11111)=1$, it follows that 99 999. .
Let $A=99999 A_{0}, A_{0} \in \mathbf{Z}_{+}$.
Since $10^{9}\frac{10^{9}}{10^{5}}=10^{4}, 10^{5}+10, a_{i} \in\{0,1, \cdots, 9\},(i=0,1,2,3,4)$, then $A=A_{0}\left(10^{5}-1\right)$
$$
=\overline{a_{0} a_{1} a_{2} a_{3} a_{4} 00000}-\overline{a_{0} a_{1} a_{2} a_{3} a_{4}} \text {. }
$$
$$
\text { Let } A=\overline{c_{0} c_{1} c_{2} c_{3} c_{4} c_{0}^{\prime} c_{1}^{\prime} c_{2}^{\prime} c_{3}^{\prime} c_{4}^{\prime}} \quad\left(c_{0} \neq 0\right) \text {. }
$$

If $a_{4} \neq 0$, then $c_{0}=a_{0}, c_{1}=a_{1}, c_{2}=a_{2}, c_{3}=a_{3}$,
$$
\begin{array}{l}
c_{4}=a_{4}-1, c_{0}^{\prime}=9-a_{0}, c_{1}^{\prime}=9-a_{1}, c_{2}^{\prime}=9-a_{2}, c_{3}^{\prime}= \\
9-a_{3}, c_{4}^{\prime}=10-a_{4} ;
\end{array}
$$

If $a_{4}=0$, then $a_{3} \neq 0$ (otherwise $A$ would have two
$$
\begin{array}{l}
9) \text {. Thus, } c_{0}=a_{0}, c_{1}=a_{1}, c_{2}=a_{2}, c_{3}=a_{3}-1, c_{4}=9, \\
c_{0}^{\prime}=9-a_{0}, c_{1}^{\prime}=9-a_{1}, c_{2}^{\prime}=9-a_{2}, c_{3}^{\prime}=10-a_{3}, \\
c_{4}^{\prime}=0 .
\end{array}
$$

Therefore, $c_{i}+c_{i}^{\prime}=9(0 \leqslant i \leqslant 4$, and $i \in \mathbf{N})$.
$$
\begin{array}{l}
\text { Hence, }\left\{\left(c_{i}, c_{i}^{\prime}\right) \mid 0 \leqslant i \leqslant 4, i \in \mathbf{N}\right. \\
\quad=\{(0,9),(1,8),(2,7),(3,6),(4,5),(9,0), \\
(8,1),(7,2),(6,3),(5,4)\} .
\end{array}
$$

Since $c_{0} \neq 0$, the number of $A$ satisfying condition (1) is $5!\times 2^{5}-4!\times 2^{4}=3456$.
Conversely, any number satisfying condition (1) can be expressed as:
$$
\begin{array}{l}
\overline{c_{0} c_{1} c_{2} c_{3} c_{4} 00000}+\overline{c_{0}^{\prime} c_{1}^{\prime} c_{2}^{\prime} c_{3}^{\prime} c_{4}^{\prime}} \\
=\overline{c_{0} c_{1} c_{2} c_{3} c_{4}} \cdot\left(10^{5}-1\right)+99999 \\
\end{array}
$$

which is necessarily divisible by 11111.
Therefore, $|M|=3456$.