 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Niners (Posted on 2002-10-23) 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.

 See The Solution Submitted by levik Rating: 4.2500 (8 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(3): | Comment 6 of 15 | 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(j-i)*(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(j-i) is divisible by n and you've found your multiple
 Posted by TomM on 2002-10-24 01:32:40 Please log in:

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