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        riddles >> easy >> Squares to zero
        
(Message started by: pcbouhid on Dec 12^th, 2005, 5:02pm)

Title: Squares to zero
Post by pcbouhid on Dec 12^th, 2005, 5:02pm

Add only (+) and (-) signs between 1^2, 2^2, 3^2... and 2005^2 and make it finally equal zero.

Title: Re: SQUARES TO ZERO
Post by Eigenray on Dec 12^th, 2005, 10:18pm

Can't be done. [hide]Any such sum will always be odd[/hide]. Unless you had something else in mind?

Title: Re: SQUARES TO ZERO
Post by pcbouhid on Dec 13^th, 2005, 9:21am

on 12/12/05 at 22:18:50, Eigenray wrote:

Can't be done. [hide]Any such sum will always be odd[/hide].

Right! Now, make it equal to 1.

Title: Re: SQUARES TO ZERO
Post by Eigenray on Dec 13^th, 2005, 11:59am

[hideb]First note that since -n²+(n+1)²=(2n+1), we can replace a pair n²,(n+1)² by 2n+1. Furthermore, since
n² - (n+1)² - (n+2)² + (n+3)² = -(2n+1) + (2n+5) = 4,
we can replace any block of four consecutive squares by a 4. Thus we can reduce to the numbers
1, 5, 9, 13, 17, 4,4,4,...,4 (499 4's),
where each of 5,9,13,17 represents 2 squares of the original, and each 4 is a block of 4 squares. But
1 = 1 + 5 + 9 + 13 + 17 + 4*244 - 4*255,
which corresponds the the sequence of signs
+, -+, -+, -+, -+, [+--+, +--+, ... (244 times)], [-++-, -++-, ... (255 times)].[/hideb]

Title: Re: SQUARES TO ZERO
Post by Barukh on Dec 14^th, 2005, 10:29am

Another arrangment would be:

+-+ (+-) [375] -- (+-) [248] ++ (+-) [375] -+