After seeing the following puzzle floating around the net, I decided to automate it:
Take the digits 9, 8, 7, 6, 5, 4, 3, 2, 1 in that order. Insert one or more basic math operators (addition, subtraction, division, multiplication) so that the result makes 2006. 
#!/usr/bin/perl

use strict;
use warnings;

my @o    = (" + ", " - ", " * ", " / ", "");
my @s    = (" ", "-");
my @d    = (0, 1 .. 9);
my $year = @ARGV ? shift : 2006;

for my $o8 (@o) {
for my $o7 (@o) {
for my $o6 (@o) {
for my $o5 (@o) {
for my $o4 (@o) {
for my $o3 (@o) {
for my $o2 (@o) {
for my $o1 (@o) {
for my $s  (@s) {
    my $expr = "$s$d[9]$o8$d[8]$o7$d[7]$o6$d[6]$o5" .
                 "$d[5]$o4$d[4]$o3$d[3]$o2$d[2]$o1$d[1]";
    print "$expr == $year\n" if $year == eval $expr;
}}}}}}}}}


__END__
2006 == -9 + 8 * 7 + 654 * 3 - 2 - 1
2006 ==  9 + 8 * 7 + 654 * 3 - 21
2006 ==  9 + 8 * 7 * 6 * 5 - 4 + 321
2006 ==  9 * 8 - 7 + 654 * 3 - 21

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I'm sure there are cleverer ways of writing the nested loop, using some kind of module. But cut-and-paste is fast, and this takes less programmer time.
Perl --((8:>*

In reply to Happy 2006 by Perl Mouse

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