Re: Diophantic Equations with Constraints

*To*: mathgroup at smc.vnet.net*Subject*: [mg49462] Re: Diophantic Equations with Constraints*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>*Date*: Wed, 21 Jul 2004 06:39:22 -0400 (EDT)*References*: <cdj1t2$nbi$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

"Michael S." <MikeSuesserott at t-online.de> wrote: > given an equation like > > 3x + 2y - z == 14 > with > x in Range[3,8], > y in Range[0,12], > z in Range[1,9], > > what would be the best way to solve this? In[1]:= Reduce[{3*x + 2*y - z == 14, 3 <= x <= 8, 0 <= y <= 12, 1 <= z <= 9}, {x, y, z}, Integers] Out[1]= y == 0 && z == 1 && x == 5 || y == 0 && z == 4 && x == 6 || y == 0 && z == 7 && x == 7 || y == 1 && z == 3 && x == 5 || y == 1 && z == 6 && x == 6 || y == 1 && z == 9 && x == 7 || y == 2 && z == 2 && x == 4 || y == 2 && z == 5 && x == 5 || y == 2 && z == 8 && x == 6 || y == 3 && z == 1 && x == 3 || y == 3 && z == 4 && x == 4 || y == 3 && z == 7 && x == 5 || y == 4 && z == 3 && x == 3 || y == 4 && z == 6 && x == 4 || y == 4 && z == 9 && x == 5 || y == 5 && z == 5 && x == 3 || y == 5 && z == 8 && x == 4 || y == 6 && z == 7 && x == 3 || y == 7 && z == 9 && x == 3 David Cantrell