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)" >