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: Picky, picky! - I think I got it !! by pcbouhid)

If in the "six" group

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

I make (x) = (1-x) ===> x = 1/2, I'll obtain :

(1) (1/2) (-1/2) (-1) (-1/2) (1/2)...the next will be (1).

Transposing the (1) :

(1/2) (-1/2) (-1) (-1/2) (1/2) (1).....the next is (1/2) = the first.

So, using 332 of such "six" groups, 6 x 332 = 1992, and adding two five strings [(1/2) (-1/2) (-1) (-1/2) (1/2)], I'll add 10 more terms, being the last term equal to the first term of the 332 groups, closing the loop.

So, the answer is 1992 + 10 - 1 = 2001.  

 

Posted by pcbouhid on 2005-05-30 22:09:16