• Mathworld
• The On-Line Encyclopedia of Integer Sequences
• PARI/GP computer algebra system

Specific to this problem:

• Divisor function
• Sloane A000005
• Math::Pari

Here's my solution (runs in about 0.6 sec):

use Math::Pari qw(numdiv);

while( ++$i ) {
    $n += $i;
    $d = numdiv($n);
    last if $d > 500;
}

print "$n: $d\n";
[download]

Another site that may be of interest is Dictionary of Algorithms and Data Structures.

Also, when you say "500 divisors" do you mean 500 distinct divisors (2⁵⁰⁰ has one divisor, repeated 500 times)?

No, 2⁵⁰⁰ has 501 divisors. It's divisible by 1, 2, 4, ..., 2⁴⁹⁹, 2⁵⁰⁰.

OK; 2 distinct divisors, one of which is repeated 500 times...

---

emc

Information about American English usage here and here.

Any New York City or Connecticut area jobs? I'm currently unemployed.

Also, when you say "500 divisors" do you mean 500 distinct divisors?

I believe that's a correct observation.
--
Andreas

I try my best at Project Euler. Both fun and highly challenging.

Problem 12 reads Which is the first triangle number to have over five-hundred divisors?

I also play with PE and this is what I had used:

#!/usr/bin/perl -l

use strict;
use warnings;

my $n;
my @div=(0);

while (++$n) {
    my ($m,$p)=($n,0);
    $m /= 2 and $p++ until $m & 1;
    $div[$n] = $p ?
      $div[$m]*$p :
      grep !($n%$_), 1..$n;
    print $n*($n-1)/2 and last
      if $div[$n]*$div[$n-1] > 500;
}

__END__
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sub get_nof2 {
    return scalar @{ divisors(+shift) };

}

{
my $D;
sub divisors {
    my $n = shift;
    foreach my $d ( 2 .. sqrt($n) ) {
        unless ( $n % $d ) {
            my $div = $n / $d;
            unless ( defined $D->{$div} ) {
              $D->{$div} = divisors($div); }
            return $D->{$n} =
              [ uniq( @{ $D->{$div} }, map { $d * $_ } @{ $D->{$div} }
+ ) ];
        }
    }
    return $D->{$n} = [ 1, $n ];
}
[download]