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Re: Solving equations and inequalities with Reduce - how?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg87715] Re: Solving equations and inequalities with Reduce - how?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 16 Apr 2008 05:01:30 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <fu1tuf$ogc$1@smc.vnet.net>

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.

Marc,

First, one of your variables had been set to the value one before you 
tried Reduce[].

Second, the option Modulus->9 is not doing what you may have in mind 
(the solution yield by Reduce[] does not fulfill your original criteria ).

Third, using FindInstance[] may help to formulate and check your problem 
in an algebraic way.

For instance, starting with a fresh kernel,

In[1]:= eqn =
   200000 a + 20000 b + 2000 c + 200 d + 20 e + 2 f ==
    100000 c + 10000 d + 1000 e + 100 f + 10 a + b;

In[2]:= Reduce[eqn && a != 0, {a, b, c, d, e, f}, Modulus -> 9]

Out[2]= a != 0 && a == C[1] && b == C[2] && c == C[3] && d == C[4] &&
  e == C[5] && f == 8 C[1] + 8 C[2] + 8 C[3] + 8 C[4] + 8 C[5]

In[3]:= FindInstance[eqn && a != 0, {a, b, c, d, e, f},
  Modulus -> 10]
eqn /. %[[1]]

Out[3]= {{a -> 1, b -> 0, c -> 0, d -> 0, e -> 0, f -> 0}}

Out[4]= False

In[5]:= FindInstance[eqn && a != 0, {a, b, c, d, e, f}, Integers]
eqn /. %[[1]]

Out[5]= {{a -> 1, b -> 4, c -> 0, d -> 0, e -> 0, f -> 2857}}

Out[6]= True

Regards,
-- Jean-Marc



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