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?
Best Answer
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.