The set of numbers {9, 99, 999, 9999, ...} has some interesting properties. One of these has to do with factorization. Take any number n that isn't divisible by 2 or by 5. You will be able to find at least one number in the set that is divisible by n. Furthermore, you won't need to look beyond the first n numbers in the set.

Prove it.

(*from http://www.ocf.berkeley.edu/~wwu/riddles/)*

all numbers divisible by 2 and 5 dont have a '9' at the end. All the others may have. So I guess those numbeer's multiples will eventually be one of the set's.