diff --git a/challenge-089/james-smith/perl/ch-1.pl b/challenge-089/james-smith/perl/ch-1.pl
new file mode 100644
index 0000000000..0e79706771
--- /dev/null
+++ b/challenge-089/james-smith/perl/ch-1.pl
@@ -0,0 +1,43 @@
+#!/usr/local/bin/perl
+
+use strict;
+
+use warnings;
+use feature qw(say);
+use Test::More;
+
+is( gcd_sum(1),   0 );
+is( gcd_sum(2),   1 );
+is( gcd_sum(3),   3 );
+is( gcd_sum(4),   7 );
+is( gcd_sum(5),  11 );
+is( gcd_sum(6),  20 );
+is( gcd_sum(7),  26 );
+is( gcd_sum(8),  38 );
+is( gcd_sum(9),  50 );
+is( gcd_sum(10), 67 );
+
+done_testing();
+
+sub gcd_sum {
+  my $n = shift;
+  my $sum = 2*$n-2; ### We can pull out the GCDs with 1 as either side
+                    ### Note because of the optimization below we only
+                    ### Remove -2 as we remove another one is the
+                    ### special case where $x==2;
+  foreach my $x ( 2..$n ) {
+    ## Note that the gcd is symetric gcd($m,$n) == gcd($m,$m-$n) where $n between 1 & $m-1
+    $sum += 2*gcd( $x, $_ ) foreach 2..$x/2;
+    ## We have counted the middle value twice if $x is even.... so remove one of them...
+    $sum -= $x/2 unless $x%2;
+  }
+  return $sum;
+}
+
+sub gcd {
+  my( $n,$m ) = @_;
+  ## The way we have set up gcd sum we know $n>$m in all cases
+  ## So don't need to check m will always be <= n
+  ($n,$m) = ( $m, $n % $m ) while $n % $m;
+  return $m;
+}
diff --git a/challenge-089/james-smith/perl/ch-2.pl b/challenge-089/james-smith/perl/ch-2.pl
new file mode 100644
index 0000000000..25befeef39
--- /dev/null
+++ b/challenge-089/james-smith/perl/ch-2.pl
@@ -0,0 +1,34 @@
+#!/usr/local/bin/perl
+
+use strict;
+use warnings;
+use feature qw(say);
+
+say map { "@{$_}\n" } @{$_} foreach magic();
+
+## We now $e must be 5... so we only need to loop through values 1..9
+## for both a and b with possible values such that $c < 10
+## the inner loop is then executed just 32 times....
+
+## We can generate all values if we only have a b & e
+## If we do this we get the sums of the 2nd and 3rd row as 3e & 30-3e respectively
+## For these to both be 15 e must be 5.
+
+sub magic {
+  my @solutions = ();
+  foreach my $a (1..4,6..9) { ## a can't be 5...
+                    ## $b can't be 5, can't be $a and can't make $c 5
+    foreach my $b ( grep { $_!=5 && $_!=$a && $a+$_!=10 } ($a<6?6-$a:1)..($a<6?9:14-$a) ) {
+      ## Only 32 get here!
+
+      ## Check digits unique and between 1 and 9 { all digits can be computed in terms of a & b }
+      my $digits = [ [ $a,         $b,    15-$a-$b   ],
+                     [ 20-2*$a-$b, 5,     2*$a+$b-10 ],
+                     [ $a+$b-5,    10-$b, 10-$a      ] ];
+
+      ## Convert into the required grid if numbers unique and 1..9
+      push @solutions, $digits if '123456789' eq join q(), sort map {@{$_}} @{$digits};
+    }
+  }
+  return @solutions;
+}