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?

My "six" group is :

(1) (x) (x-1) (-1) (-x) (1-x)

or, transposing the (1),

(x) (x-1) (-1) (-x) (1-x) (1)

For x = 0,

(0) (-1) (-1) (0) (1) (1)

333 such groups give us 6 x 333 = 1998 numbers, starting with a (0). Adding a group (0) (-1) (-1) (0), this last (0) will close the loop with the first (0), so I will add only 3 number, (0), (-1) and (-1).

The series is [(0) (-1) (-1) (0) (1) (1)] (0) (-1) (-1), the terms between the [] 333 times.

So, 1998 + 3 = 2001.  

Posted by pcbouhid on 2005-05-31 20:08:22