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
re(2): by Cheradenine)
I think I see C9's reasoning:
Take the first n members of set N and find the remainders when you divide by n.
There are two cases:
Case 1 (All of the members yeild different remainders): Then since there are n different remainders, and there are only n possible remainders, including 0, one of those remainders must be 0
Case 2 (Two or more have the same remainder): Take two of the numbers with the same remainder N(i) and N(j) and subtract them N(j)N(i) = N(ji)*(10^i) This is divisible by n, so either (10^i) is divisible by n (Which is why the proof does not include 2 or 5) or N(ji) is divisible by n and you've found your multiple

Posted by TomM
on 20021024 01:32:40 