That was my initial thought but that would make 30 => 2 and not 1. So I think he wants int( ($num-1) / 30 ) + 1.

#!/usr/bin/perl

use strict;
use warnings;

my @nums = ( 1 .. 1000 );

foreach my $num ( @nums ) {
   my $test_n =  int( ($num-1) / 30 ) + 1;
   print "Num: $num - $test_n \n";
}
[download]

Um, technically this won't work either for $total_nums = 1, much as friedo's version doesn't work for $total_nums values of 30, 60, 90, etc., based on the OP's original code. Looks like diotalevi's example using ceil() is better.

Update: Ah, I see you updated your version to add 1 after getting the int() value. Fine.

(Blush)Another Update: Uh, actually, not fine. Now yours doesn't work for values of 31, 61, etc.

Wha??? Sure it does 31 => 2 and 61 => 3 just like the OP:

$ perl -e 'print int( (31-1) / 30 ) + 1, "\n"'
2
$ perl -e 'print int( (61-1) / 30 ) + 1, "\n"'
3
[download]

-derby