What could be helpful for you is the modulus operator as explained in http://perldoc.perl.org/5.8.9/perlop.html#Multiplicative-Operators. $n % $p equals zero if $p is a divisor of $n.

How about some golf?

($%%$_)||print$_.$/for 1..($%=shift)
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Nice! I particularly like the $%%$_ construct. But why so verbose?

 1:49 >perl -wE"$%%$_||say for 1..($%=pop)" 36
1
2
3
4
6
9
12
18
36

 1:49 >
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Ten characters strokes saved. :-)

Athanasius <°(((>< contra mundum

Iustus alius egestas vitae, eros Piratica,

> How about some golf?

more obfuscation than golf. :)

Beside the cryptic variable-name, is the first paren not needed and you can use say to shorten print.

Cheers Rolf
_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Je suis Charlie!

First step: How would you find the divisors if you had no computer?

How can I write this program in perl String r = ""; int p = 2; while (n > 1) if (n%p == 0) { r += p+","; n /= p; } else p++;

The snippet you provided prints factorisation (i.e. only prime divisors, not all divisors). Translating it to Perl is straightforward:

#!/usr/bin/perl
use warnings;
use strict;

my $n = <>; # read the input

my $r = '';
my $p = 2;
while ($n > 1) {
    if ($n % $p == 0) {
        $r .= "$p, ";  # . is the concatenation operator
        $n /= $p;
    } else {
        ++$p;
    }
}
print "$r\n";
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لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ

What did you try? ¹

¹) What, not even googling? ;)

Like to learn to write this program in perl String r = ""; int p = 2; while (n > 1) if (n%p == 0) { r += (p+","); n /= p; } else p++;

Care to provide us with those crosslinks, so we don't waste time looking for them?

I would assume that an application of such an algorithm is a stable of introductory CS or programming courses everywhere. It would not surprise me to find (multiple) such questions in the various venues that people tend ask them.