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Topic: mod 11
Replies: 11   Last Post: May 16, 2006 11:43 AM

 Messages: [ Previous | Next ]
 Dr. Eric Wingler Posts: 139 Registered: 12/12/04
Re: mod 11
Posted: May 15, 2006 3:40 PM

"mina_world" <mina_world@hanmail.net> wrote in message
> hello.....doctor~
>
> [(1^2)/11]+[(2^2)/11]+[(3^2)/11]+....+[(2004^2)/11]
>
> = ? (mod 11) , [~] is Gauss function.
>
> find the ? .
> -------------------------------------------------

sum( [(i^2)/11], i = 1 to 2004) = sum((i^2) / 11 - r_i / 11, i = 1 to 2004),
where r_i is the quadratic residue of i^2. The sum r_1 + r_2 + . . . +
r_2002 is easy to compute because the quadratic residues are periodic with
period 11.

________________________________
Eric J. Wingler (wingler@math.ysu.edu)
Dept. of Mathematics and Statistics
Youngstown State University
One University Plaza
Youngstown, OH 44555-0001
330-941-1817