SPUG: FW: Help..Any Math GENIUS out there?
Sanford Morton
smorton at pobox.com
Thu Jul 25 16:41:00 CDT 2002
If A always has a unit coefficient, then it looks like a factor with A the
remainder. I haven't checked this:
perl -e '
$e = int (2307012 / 129600); $e1 = 2307012 % 129600;
$d = int ($e1 / 3600); $d1 = $e1 % 3600;
$c = int ($d1 / 360); $c1 = $e1 % 360;
$b = int ($c1 / 36);
$a = $c1 % 36;
print "$a $b $c $d $e \n";
'
On Thu, 25 Jul 2002, Orr, Chuck (NOC) wrote:
>
>
> > Hello,
> >
> > I am hoping to use Perl to solve an equation like the below:
> >
> > A + (36*B) + (360*C) + (3600*D) + (129600*E) = 2,307,012
> >
> > I know that each of the variables A - E is a whole number in the range
> > 1-36.
> >
> > Could I load those numbers (1 -36) in an array and substitute each number
> > for each variable until this is solved?
> >
> > Any ideas would be greatly appreciated.
> >
> >
> > Thanks,
> > Chuck Orr
> > AT&T Wireless Services
> > 425 288 2386
> > chuck.orr at attws.com
>
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