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diff --git a/challenge-149/james-smith/README.md b/challenge-149/james-smith/README.md
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+++ b/challenge-149/james-smith/README.md
@@ -20,6 +20,27 @@ https://github.com/drbaggy/perlweeklychallenge-club/tree/master/challenge-149/ja
 
 ## The solution
 
+```perl
+for( my($n,$ds,$i,$fa,$fb,%fib)=(@ARGV?$ARGV[0]:20,0,0,1,1,0,1,1,1);
+     $n; $i++,$ds=0 ) {                                 ## 1
+  $ds+=$_ foreach split //,$i;                          ## 2
+  ($fib{$fa+$fb},$fa,$fb)=(1,$fb,$fa+$fb) if $ds > $fb; ## 3
+  (say $i)**$n-- if exists $fib{$ds};                   ## 4
+}
+```
+
+**Notes:**
+
+ * Line 1 - We initialise everything inside the for loop
+   * `$n` is the number to print (and is based on what is passed at the command line)
+   * `$ds` is the digit sum (Note we reset it everytime through the loop in the incremement part of the loop
+   * `$i`  current value being considered
+   * `$fa` & `$fb` - the highest two fibonacci numbers 
+   * `%fib` hash whose keys are fibonacci numbers
+ * Line 2 - Computes the digit sum by splitting number on `//` I split into 1 character blocks
+ * Line 3 - Expand the fibonacci hash by 1 if the digit sum is greater than the highest fibonnaci number {we don't need to loop this as the digit sum of `$n+1` can only be at most 1 higher than that for `$n`. Note we just update $fb and $fa in this line
+ * Line 4 - Check to see if the digit sum exists, print and decrement counter - and return to the start of the loop.
+
 # Challenge 2 - Largest Square
 
 ***Given a number base, derive the largest perfect square with no repeated digits and return it as a string. (For base>10, use ‘A’..‘Z’.)***