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?

(In reply to re(5): What now ? - you're right again !! by Charlie)

I agree with you, Charlie. In fact, I already did.

But, McWorter is "joking" with us ? I don't think so, once it was proved the "six" cycle.

I still think that there's something we are missing, because the solution of 2004 is too obvious. 

Posted by pcbouhid on 2005-06-01 16:40:41