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Re: Russian Peasant Multiplication / was question on

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  • Subject: [mg131408] Re: Russian Peasant Multiplication / was question on
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Wed, 3 Jul 2013 05:02:26 -0400 (EDT)
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x = RandomInteger[{-1000000, 1000000}]


99333


id = IntegerDigits[x]


{9, 9, 3, 3, 3}


pwr10 = 10^Range[Length[id] - 1, 0, -1]


{10000, 1000, 100, 10, 1}


id*pwr10


{90000, 9000, 300, 30, 3}


x ==
 Sign[x]*id.pwr10 ==
 Sign[x]*Dot[id, pwr10] ==
 Sign[x]*Total[id*pwr10] ==
 Sign[x]*Plus @@ (id*pwr10)


True



Bob Hanlon




On Tue, Jul 2, 2013 at 12:49 AM, <d.a.paxton at gmail.com> wrote:

> On Saturday, June 29, 2013 2:47:03 AM UTC-6, Richard Fateman wrote:
> > Dave --
> >
> > 1.  You should try to come up with a useful subject line
> >
> > in the future.
> >
> > 2. It is called Russian Peasant Multiplication (which you may
> >
> > find on Google).
> >
> > 3. There is no reason to believe that a procedural algorithm
> >
> > has a formula, but in this case I think the inverse is known
> >
> > as division :)
>
> Bob,  Thanks on that.  I see the use of the floor function.  This whole
> idea is trivial when one can use that.  What I am trying to do is basically
> do the floor function as a linear equation.  This is useful for pulling
> numbers apart.  Like a number 13.  How does one algebraically make it 1 and
> 3 or 10 and 3.  Either.
>
>




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