Prove that there is a multiple of 2009 that ends with the digits 000001.
May one generalise this to: There exists a multiple of $x$ that ends with the digits $y$ (where $y$ consists of $n$ digits) if $x$ and $10^n$ are relatively prime?
The proof may be constructive, or non constructive.
Any help would be greatly appreciated.
I figured the multiple has to end in a $9$: $2009 \cdot 889$ gives $001$ as the three ending digits. I then tried to see if I could get the $4$th last digit to be a zero, but I must admit I am clearly missing something at that point.