Couldn't sleep, so I put some stuff I had together: 

#!/usr/bin/perl -l

# http://perlmonks.org/?node_id=1169564

use strict;

$_ = 6652;

1 while do {
  /\d.*?\d(?{ 17 == eval "$`($&)$'" and print "$`($&)$'" })(*FAIL)/
    for map tr#a-d#-+*/#r, glob join '{a,b,c,d}', split //
  },
  s/.*\K # find the last
  (.) # digit such that
  (.*) # there is a later (latest)
  (.)(??{$1 <= $3 and '(*FAIL)'}) # digit less than it
  (.*) # and get rest
  # swap those two digits ( $1 & $3 )
  # then reverse everything after the first swapped digit
  / $3 . reverse $2.$1.$4 /xe;

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In reply to Re: Solution to A simple but difficult arithmetic puzzle by Anonymous Monk
in thread Solution to A simple but difficult arithmetic puzzle by talexb

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