Four, among the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion.

In the 10-digit numbers where each digit is different, how many are multiples of 11111? Prove your conclusion.
Let $n=\overline{a b c d e f g h i j}$ satisfy the conditions, then $11111 \mid n$, and $a, b, \cdots, j$ are a permutation of $0,1,2, \cdots, 9$ $(a \neq 0)$.
Since $a+b+\cdots+j=0+1+\cdots+9=45$,
it follows that $9 \mid(a+b+\cdots+j)$. Therefore, $9 \mid n$.
Since 11111 and 9 are coprime, then $99999 \mid n$.
Let $x=\overline{a b c d e}, y=\overline{f g h i j}$, then $n=10^{5} x+y \equiv x+y \equiv 0(\bmod 99999)$.
But $0<x+y<2 \times 99999$, so it must be that $x+y=99999$,

which means $a+f=b+g=c+h=d+i=e+j=9$.
Allowing $a=0$ first, then $(a, f),(b, g),(c, h),(d, i)$, $(e, j)$ are a permutation of $(0,9),(1,8),(2,7),(3,6),(4,5)$, and the two numbers in each pair can be swapped, hence there are $5! \times 2^{5}$ possibilities.
When $a=0$, $f=9$, in this case, there are $4! \times 2^{4}$ possibilities:
In summary, the number of 10-digit numbers that satisfy the conditions is
$$
5! \times 2^{5} - 4! \times 2^{4} = 9 \times 4! \times 2^{4} = 3456 \text{. }
$$