Counting ten-digit numbers whose digits are all different and that are divisible by $11111$ I've got a problem which I can't finish off and I've been trying for long long time without sucess. I think it's easier than I thought but, maybe yes o maybe not nevertheless I'm trying to do by myself.
The problem says

Call a number interesting if is a $10$-digit number, all of its digits are different and is divisible by $11111$. How many interesting numbers are there?

Hitherto, I've tried things like cheking the numbers $\equiv (mod 
 9)$ and $ \equiv (mod 100000) $  and also have checked a base-10 decomposition and I suspecting that there are just 1 or 2 number with the desire properties or there aren't. Should I proced by contradicition? And if I would, any idea?
 A: The comments indicate that since $9\nmid11111$ yet the sum of digits of an interesting number is divisible by $45$, all interesting numbers are divisible by $99999$; it is then not hard to show that in such numbers the last five digits are complements ($x\to9-x$) of the first five.
The complementary pairs are $09,18,27,36,45$, so there are $5!\cdot2^5$ ways of assigning pairs to the first five positions of an interesting number and then choosing explicitly the digits that go there – except that $4!\cdot2^4$ must be subtracted for those choices giving a leading zero. This leaves $3456$ interesting numbers.
A: Another way to obtain the same is the following. We have to count the number of ways to fill first $5$ digits of the number only, the second half is (digit-wise) complement of it and is uniquely determined by the first half. So in the following $10$-digit number
$$ABCDE - A'B'C'D'E'$$
there are $9$ ways to pick $A$ (excluding $0$), $8$ ways to choose $B$ (since $A$ and $A'$ are filled), $6$ ways for $C$, $4$ for $D$ and $2$ for $E$. Required number is
$$9 \times 8 \times 6 \times 4 \times 2 = 3456$$
