noonsun

So...?

I'll just call it r then.

So if you make a list of 2009 numbers that goes like this:
12
1212
121212
...
121212...1212

you get 2009 r-s if a is one of those numbers and b is 2009. If any of those r-s is 0, then the solution is to add as much zeros as you want on the number 12...12 whose r is 0 (because if r=0 that means that the number is divisible with 2009, and if you add zeros it will still be divisible with 2009 and it will look like this: 121212...121200...000). If none of the r-s is 0 then you can have maximum 2008 different r-s, and because you have 2009 numbers, at least 2 of the r-s are same. So let the first 12...12 number be k and the second l (the numbers with the same r-s). The solution is k-l. I hope you get it.

I just need you to translate something for me first. How do you call the r part in a=b*k+r ? (That's a way you can write any a using b, like 13=5*2+3)

Awesome.
Like really, really awesome.

Would you know how to prove that there is a number like 121212...1212000...00 divisible with 2009?

Hello.

Great!

Hi! How are you?

...He actually hasn't said anything mean to me in a while... hmmm... is he sick???

but i was being serious