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MathGroup Archive 2005

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Re: ChineseRemainder

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57077] Re: ChineseRemainder
  • From: "Dana DeLouis" <delouis at bellsouth.net>
  • Date: Sat, 14 May 2005 04:58:19 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Thank you everyone.  I see now that when the interval of list2 falls within
the interval of list1, it is possible to get a returned value 'R' such that
Mod[R, list2] will "not" match list1.  I'm still doing some study on my own
with my "number theory" book in hand.  Thanks again.  :>)
-- 
Dana

 

 

"Dana DeLouis" <delouis at bellsouth.net> wrote in message
news:d5ski6$nvj$1 at smc.vnet.net...

> Hello.  I have a question on the function "ChineseRemainder."  Could
anyone
> offer an explanation on the following behavior?  Thanks.
> 
 Here's the package...
 Needs["NumberTheory`NumberTheoryFunctions`"]
<skip.

 

> list1={2,3,4};list2={9,7,4};
> 
> r=ChineseRemainder[list1,list2]
> 
> 164
> 
> However, it does not correctly return list1.  (According to help)
> 
> Mod[r,list2]
> 
> {2,3,0}
> 
> This happens often in my program, and I'm having a tough time trusting the
> solution.  Does anyone familiar with this have any insight?  Thanks.
> 
> I note that Mathematica's built-in ChineseRemainder function returns the
> same answer, so I'm sure Mathematica is doing it correctly.  I guess I
don't
> understand why Mod[r,list2] will often not return list1 (According to
help).
> 
> Reduce`RChineseRemainder[list1, list2]
> 
> 164
> 
> Thank you for any insight.
> Dana
> 
> $Version
> "5.1 for Microsoft Windows (January 27, 2005)"
>




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