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Re: numerical challenge
Posted:
Oct 23, 2012 8:25 PM
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"Sylvia Else" <sylvia@not.at.this.address> wrote in message
news:aeolaqFm67lU1@mid.individual.net...
> On 23/10/2012 12:31 PM, RichD wrote:
> > I saw this posed as an interview question:
> >
> > (usual exponent notation)
> > What are the last 3 digits of 171 ^ 172?
> >
> > Presumably, one is given pen and paper.
> >
> > Is there a trick here? The second digit isn't too tough,
> > but the third is a lot of work.
> >
> > --
> > Rich
> >
>
> This appears to originate from
>
>
http://blog.brilliantscholars.com/2012/10/08/live-challenge-algebranumber-th
eory-5/
>
> Solutions similar to those that I and others have posted in this thread
> are given there, but the original post implies that there's a single
> line solution. If there is, I can't see it.
>
> I have noted that 171 = 900 - 9^3, which looks a bit interesting, but I
> haven't been able to make anything of it in the sense of achieving a
> one-line answer.
>
> Sylvia.
Well, perhaps: (dropping all multiples of 1000 as in your other posts)
( 900 - 9^3 )^172 = ( 100 + (10-1)^3 )^172
= (10 - 1)^516 + 100 * 172 * (10 - 1)^513
= (1 - 160) + 200 * (-1 + 130)
= 1 - 160 - 200 = 641
This is easy on the calculations, but not just a single line....
Regards,
Mike.
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