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MathGroup Archive 2005

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ChineseRemainder

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56971] ChineseRemainder
  • From: "Dana DeLouis" <delouis at bellsouth.net>
  • Date: Wed, 11 May 2005 05:25:04 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

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`"]

The following small example has no solution, and returns the null set, which
is ok...

list1={2,3,4};list2={9,4,8};

r=ChineseRemainder[list1,list2]

{}


The following returns a solution of 94.

list1={2,3,4};list2={4,7,9};

r=ChineseRemainder[list1,list2]

94

According to Help on this function, we can test the solution with the
following, and it correctly returns list1.

Mod[r,list2]

{2,3,4}


These similar numbers return a different solution.

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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