Over 2000 numbers are around a circle. Each number is the sum of its left and right neighbors.

Given that one of the numbers is a one, how many numbers (as a minimum) must there be?

Though I (still, until now) agree with the answer 2004, I don't think that the proof must exaust the analysis only for 0, 1, 2, 3, 4 or 5 cycles.

Just as an example (not that it is possible) : 2001 = 3 x 23 x 29.

If we could find a cycle of 23 numbers, then 89 such cycles will give us 2001 numbers. I tried to find a 23-cycle, using the 6-cycle, using (x) as a term, and trying to achieve the value of x that could close a 23-cycle (4x6-1), but I didn't succeed.

Yet, 29 = (5x6)-1. I didn't try to achieve a loop of 29 (69 groups of them would lead also to 2001 numbers).

The same reasoning could be made with 2002 = 2 x 7 x 11 x 13. But I confess that the way that I'm trying is too tiring. There's must be a more simple way.

 

Posted by pcbouhid on 2005-06-02 19:03:52