Re: Math Question....

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On Thu, Apr 22, 2010 at 10:29 AM, Richard Quadling <rquadling@xxxxxxxxxxxxxx
> wrote:

>  >
> > It sounds like you are looking for factors.
> >
> >
> http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver
> >
> > Solution by Find factors of any number
> >
> > 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819
> > Work Shown
> >
> > 1252398 is divisible by 2: 1252398 = 626199 * 2.
> > 626199 is divisible by 3: 626199 = 208733 * 3.
> > 208733 is divisible by 7: 208733 = 29819 * 7.
> > 29819 is not divisible by anything.
> >
> > So 29819 by 42 (7*3*2)
> >
> > would be a route.
>
> Aha. Missed the "30" bit.
>
> So, having found the factors, you would need to process them to find
> the largest combination under 30.
>
> 2*3
> 2*3*7
> 2*7
> 3*7
>
> are the possibilities (ignoring any number over 30).
>
> Of which 3*7 is the largest.
>
> So, 1,252,398 divided by 21 = 59,638
>
>
> Is that the sort of thing you are looking for?
>
>

Yes, that looks exactly what like what I'm looking for.  I'm going to try
and wake up the algebra side of my brain that hasn't been used in years and
see if I can digest all this.

For the 2, 3, and 7, that is based solely on the last number being divisible
by a prime number?

Joao, Jason, thanks for the code.

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