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)"

**Follow-Ups**:**Re: ChineseRemainder***From:*Chris Chiasson <chris.chiasson@gmail.com>