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 quick answer by armando)

The proof may not be completely wrong. The first loop has six numbers, but the next loop shares the last number of the first loop...doesn't that make the next loop only 5 numbers....making the final loop sharing the first and last numbers? Would that make it a different number than a multiple of six?

Posted by Brad on 2005-05-27 17:38:30