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Re: Diophantic Equations with Constraints
*To*: mathgroup at smc.vnet.net
*Subject*: [mg49457] Re: [mg49444] Diophantic Equations with Constraints
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 21 Jul 2004 06:39:17 -0400 (EDT)
*References*: <200407201153.HAA23713@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 20 Jul 2004, at 20:53, Michael S. wrote:
> *This message was transferred with a trial version of CommuniGate(tm)
> Pro*
> The equation should read
>
> 3x + 2y - z == 14 (not 148)
>
> Anyway, it's the principle of the thing I'd be interested in.
>
> Thanks,
>
> Michael
>
With the new equation we get:
Reduce[{3*x + 2*y - z == 14, 3 <= x <= 8, 0 <= y <= 12,
1 <= z <= 9}, {x, y, z}, Integers]
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
and
NMinimize[{Abs[3*x + 2*y - z - 14],
3 <= x <= 8 && 0 <= y <= 12 && 1 <= z <= 9 &&
(x | y | z) $B":(B Integers}, {x, y, z}]
{0., {x -> 5, y -> 0, z -> 1}}
NMinimize normally just finds one solution and it turns out to be the
first one returned by Reduce.
Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/
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