Re: modulo solving lacking domain?
- To: mathgroup at smc.vnet.net
- Subject: [mg126844] Re: modulo solving lacking domain?
- From: Dana DeLouis <dana01 at me.com>
- Date: Wed, 13 Jun 2012 04:54:11 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
> Solve[12*n==8,n,Modulus->20]
> {{n->4+5*C[1]}}
>
> It omits C[1] element of Integers.
Hi. Don't know where one would find this, so try:
Solve[12*n==8,n,Integers,Modulus->20]
Solve::mdom: Warning: Mathematica is ignoring domain specification Integers;
option setting Modulus -> 20 implies domain integers modulo 20.
There you go!
.. Modulus -> 20 implies domain integers :>)
// And the solution appears to check ok.
Table[Mod[12(4+5 c), 20], {c, -3, 10}]
{8,8,8,8,8,8,8,8,8,8,8,8,8,8}
= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Math 8
= = = = = = = = = =
On Jun 12, 3:02 am, Richard Fateman <fate... at cs.berkeley.edu> wrote:
> Solve[12*n==8,n,Modulus->20]
>
> returns
> {{n->4+5*C[1]}}
>
> It omits C[1] element of Integers.
> I doubt that this is a feature; is it a bug?
>
> C[1] is not necessarily a member of the finite field of
> integers modulo 20. It is obvious not an arbitrary Real.