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(4): What now ? - you're right again !! by pcbouhid)

But it is proven that such a cycle must be a multiple of 6.  Each successive number around the circle is the previous number minus the one before that, so that starting with any two numbers, a and b, the remaining numbers are forced:

a, b, b-a, -a, -b, a-b, ...

The next number is forced to be (a-b) - (-b) = a; then comes a - (a-b) = b, so after the 6 items shown, the same 6 appear again starting with a and b.  When it comes full circle, it has to match, and so can only come in again at a multiple of 6.

The only way to have a cycle of three would be if a = -a and b = -b; but that's true only when a = b = 0.

The only way to have a cycle of two would be if a = b-a = -b; but in that instance 2*b would equal -b, and b would be zero. A similar algebra on the other three terms of the 6-cycle shows a would also have to be zero, so again it would be all zeroes.

Posted by Charlie on 2005-06-01 15:20:09