use Perl

• Is this it? (Score:2)

  by AndyArmstrong (7200)

  #!/usr/bin/env perl
  
  my @candidate = grep { $_ % 10 } 101 .. 999;
  for my $N ( 7, 9, 11 ) {
      print join( ', ',
          grep { $_ % $N == 0 && reverse( $_ ) % $N == 0 } @candidate ),
        "\n";
  }

  That's my interpretation of the spec - but that's a hell of a lot more than six numbers as you say. Is that what you get?
  • Re:Is this it? (Score:2)

    by Ovid (2709) on 2008.06.08 13:07 (#63223) Homepage Journal

    Nope, not even close :) That tiny spec bears close rereading. For example, "using each of the digits 1 to 9" for 3-digit numbers means your first grep is off.

    my @candidates = grep { !/0/ } 111 ... 999;

    I also used &List::MoreUtils::uniq to pre-trim that list (since duplicate numbers are not allowed).

    Basically, you'll need a three stage process (I think). First, generate your candidate list. Second, find all three digit numbers which satisfy the reverse divisor requirement. Then construct your final list of three triples.

    And though I thought I had matched dakkar's results above, I actually have three numbers for 9 :(

    Reply to This

    Parent

    • Re: (Score:1)

      by dakkar (5832)

      Argh! *three numbers*, not *two pairs*. Stupid typing-faster-than-reading...