For the leading zero problem, use sprintf:

$formatted_date = sprintf "%.2d.%2d.%2d", $mon+1, $mday, $year%100;
[download]

Update: everything from this point on, except the very last sentence, is intended mainly for humor value. For practical advice, go straight to the end.

For the other issue, just convert your day,month,year into a number of days since a fixed date, e.g.:

=head2 greg2rd

    $rd = greg2rd( $year, $month, $day );

    Convert gregorian year,month,day to days on or after Jan 1, 1 CE
    (Gregorian).  Normalization is performed (e.g. month of 28 means
    April two years after given year) for month < 1 or > 12 or day < 1
    or > last day of month.

=cut

sub greg2rd {
    use integer;
    my ( $y, $m, $d ) = @_;
    my $adj;

    # make month in range 3..14 (treat Jan & Feb as months 13..14 of p
+rev year)
    if ( $m <= 2 ) {
        $y -= ( $adj = ( 14 - $m ) / 12 );
        $m += 12 * $adj;
    } elsif ( $m > 14 ) {
        $y += ( $adj = ( $m - 3 ) / 12 );
        $m -= 12 * $adj;
    }

    # make year positive (oh, for a use integer 'sane_div'!)
    if ( $y < 0 ) {
        $d -= 146097 * ( $adj = ( 399 - $y ) / 400 );
        $y += 400 * $adj;
    }

    # add: day of month, days of previous 0-11 month period that began
+ w/March,
    # days of previous 0-399 year period that began w/March of a 400-m
+ultiple
    # year, days of any 400-year periods before that, and -306 days to
+ adjust
    # from Mar 1, year 0-relative to Jan 1, year 1-relative (whew)

    $d += ( $m * 367 - 1094 ) / 12 + $y % 100 * 1461 / 4 +
      ( $y / 100 * 36524 + $y / 400 ) - 306;
}

# is date $formatted_then more than $autodelete days ago?
($mon,$mday,$year) = (localtime)[4,3,5];
$now = greg2rd($year+1900, $mon+1, $day);
($mon,$mday,$year) = split /\./, $formatted_then;
$then = greg2rd($year+1900, $mon, $day);
delete if ($now - $then > $autodelete);
[download]

Or instead of going to all of that bother, learn one of the fine modules that handle dates for you: DateTime,Date::Calc,Date::Manip.

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