Re: ChineseRemainder

• To: mathgroup at smc.vnet.net
• Subject: [mg56993] Re: [mg56971] ChineseRemainder
• From: Chris Chiasson <chris.chiasson at gmail.com>
• Date: Thu, 12 May 2005 02:32:32 -0400 (EDT)
• References: <200505110925.FAA24094@smc.vnet.net>
• Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
• Sender: owner-wri-mathgroup at wolfram.com

```Mod[4,4] is zero.

I think the routine should return a null set.

On 5/11/05, Dana DeLouis <delouis at bellsouth.net> wrote:
> 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)"
>
>

--
Chris Chiasson
http://chrischiasson.com/
1 (810) 265-3161

```

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