At 10:17 AM -0400 4/22/10, Dan Joseph wrote:

On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephe...@rogers.com> wrote:1,252,398 DIV 30 = 41,746 groups of 30. 1,252,398 MOD 30 = 18 items in last groupWell, 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. -- -Dan Joseph

_Dan:

`As I see it -- you are asking is "What would be the 'optimum' group`

`size for 1,252,398?"`

"Optimum" here meaning: 1. A group size of 30 or under; 2. All groups being of equal size. Is that correct?

`There may not be an exact solution, but a first order attempt would`

`be to divide the total number by 30 and check the remainder (i.e.,`

`MOD), the do the same for 29, 28, 27... and so on.`

`The group size "solution" would be a number with a zero remainder OR`

`with a remainder closest to your group size.`

That would be my first blush solution. Cheers, tedd -- ------- http://sperling.com http://ancientstones.com http://earthstones.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php