use Perl

• I bet New Scientist publishes a correction (Score:1)

  by Limbic Region (3985)

  There are a total of 24 solutions to the puzzle. These 24 solutions are comprised of 10 unique integers. If you do not consider the reverse of an integer unique, there are 5 unique integers. No matter which way you slice this - there is no way to get to "six integers" unless the spec is incomplete.
  • Re: (Score:1)

    by depesz (7751)

    24 solutions? how did get them? can you show them?
    • Re:I bet New Scientist publishes a correction (Score:1)

      by Limbic Region (3985) on 2008.06.09 7:24 (#63249) Homepage Journal

      Well, first the math
      
      there is 1 solution for 7, with reverse = 2
      there is 3 solutions for 9, with reverse = 6
      there is 1 solution for 11, with reverse = 2
      
      2 * 6 * 2 = 24
      
      I used a bit more complicated code than I am about to show, but you should be able to see how I came up with the 24 solutions

      for my $a (1 .. 9) {
          for my $b (grep {! /$a/} 1 .. 9) {
              for my $c (grep {! /$a|$b/ 1 .. 9) {
                  my $first = join '', $a, $b, $c;
                  next if $first % 7 || (scalar reverse $first) %7;
                  #....
                  print "$first $second $third"

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      • Re: (Score:1)

        by depesz (7751)

        well, you said: 1 solution for 7, with reverse 2. actually - there are 2 solutions for 7 (4 with reverse).
        
        and your code shows them.
        
        also - you didn't take into consideration the fact that digits cannot be reused between numbers generated for various dividers.
        
        
        • Huh? (Score:1)

          by Limbic Region (3985)

          With regards to the math: What I posted about 24 solutions comprised of 10 different integers was correct (as is my code). I made a mistake when I was explaining where I came up with the 24 solutions because I didn't have the code or the results in front of me. My apologies.
          
          also - you didn't take into consideration the fact that digits cannot be reused between numbers generated for various dividers
          
          I am not sure I understand.

          for my $a (1 .. 9) { # 1 - 9
              for my $b (grep {! /$a/} 1 .. 9) { # 1