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Re: Solving equations and inequalities with Reduce - how?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg87699] Re: [mg87655] Solving equations and inequalities with Reduce - how?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 16 Apr 2008 04:58:28 -0400 (EDT)
*References*: <200804150950.FAA24909@smc.vnet.net>
On 15 Apr 2008, at 18:50, Marc Heusser wrote:
> I tried to solve equations with Reduce and somehow did not quite
> formulate it right, so Reduce complains with
> "Reduce::ivar: 1 is not a valid variable".
>
> This is what I tried:
>
> Wanted: A six digit number satisfying the following conditions:
> The first digit is not zero.
> If you take the first two digits and move them to the end of the
> number,
> the resulting number must be twice the original number.
>
> In[23]:=Reduce[200000a +20000 b +2000 c+200 d+20 e + 2 f
> \[Equal]100000c +10000 d +1000 e+100 f+10 a + b , {a,b,c,d,e,f},
> Modulus\[Rule]9]
>
> I did solve the problem through exhaustive search:
>
> In[14]:=Timing[Select[Range[10^6], FractionalPart[#/10000]
> 1000000 +
> IntegerPart[#/10000]\[Equal] 2 #&]]
> Out[14]={34.2848 Second,{142857,285714,428571}}
>
> but would like to understand how to use Reduce (or another function)
> to
> solve such a set of equations.
>
> TIA
>
> Marc
>
> --
> remove bye and from mercial to get valid e-mail
> <http://www.heusser.com>
Reduce[200000*a + 20000*b + 2000*c + 200*d + 20*e + 2*f ==
100000*c + 10000*d + 1000*e + 100*f + 10*a + b &&
Inequality[1, Less, a, LessEqual, 9] && 0 <= b <= 9 && 0 <= c <=
9 &&
0 <= d <= 9 && 0 <= e <= 9 && 0 <= f <= 9, {a, b, c, d, e, f},
Integers]
(a == 2 && b == 8 && c == 5 && d == 7 && e == 1 &&
f == 4) || (a == 4 && b == 2 && c == 8 && d == 5 &&
e == 7 && f == 1)
Andrzej Kozlowski
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