On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: > On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d@xxxxxxxxxx> 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. > How do you mean break it equally? If the number doesn't fit, then you've got a remainder, and no math is going to change that. How do you want that remainder distributed? Thanks, Ash http://www.ashleysheridan.co.uk
