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/)
(In reply to
by Cheradenine)
If your reasoning were true, would this pigeonholelike principle not also mean that you could find multiples of 2 and 5 in the series?
Say that for 2, the only possible remainders are 0 and 1, but all of the numbers in the series will yield the remainder of 0.

Posted by levik
on 20021023 17:27:24 