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/)

I don't know if this is of any importance...
so the series is obviously 9 * {1, 11, 111, ... }

Which means all the elements are divisible by 3 and 9.

the third element is also divisible by 27, the 9th by 81, etc. So, the powers of 3 are covered.

at this point, the thought stalls.

Posted by levik on 2002-10-23 04:56:42