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Circle of numbers (Posted on 2005-05-27) Difficulty: 2 of 5
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?

See The Solution Submitted by McWorter    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: Picky, picky! - I think I got it !! | Comment 11 of 31 |
(In reply to Picky, picky! by McWorter)

In my previous comment, I found a string with with 6 terms :

(1) (x) (x-1) (-1) (-x) (1-x)...the next is (1).

With 333 groups of this, I could make a string of 1998, no matter the value of x.

The third term equal to 1, only add two more terms to the loop.

But if the fifth term (-x) = (1) ===> x = -1, I can add only five more terms, and close the loop with the first five terms of the group above, where I have to make x = -1.

So, 1998 + 5 = 2003 terms.


  Posted by pcbouhid on 2005-05-30 21:22:56
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