Re: Math Question....

On Thu, 22 Apr 2010 10:17:10 -0400 Dan Joseph <dmjoseph@gmail.com> wrote: > On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d@rogers.com> > wrote: > > > 1,252,398 DIV 30 = 41,746 groups of 30. > > > > 1,252,398 MOD 30 = 18 items in last group > > > Well, the only problem with going that route, is the one group is not > equally sized to the others. 18 is ok for a group in this instance, > but if it was a remainder of only 1 or 2, there would be an issue. > Which is where I come to looking for a the right method to break it > equally. > My take on it: $Items=1252398; $MaxInGroup=30; for ($x=$MaxInGroup; $x>1;$x--) { $remainder=$Items % $x; // Change 17 to the max amount allowed in the last group if ($remainder == 0 || $remainder >= 17) { // $groups = (int) ($Items /$x)+1; echo $groups."\n"; echo $remainder; break; } } -- Peter van der Does GPG key: E77E8E98 IRC: Ganseki on irc.freenode.net Twitter: @petervanderdoes WordPress Plugin Developer Blog: http://blog.avirtualhome.com Forums: http://forums.avirtualhome.com Twitter: @avhsoftware