use Perl

• Is this it? (Score:2)

  by AndyArmstrong (7200) on 2008.06.08 12:54 (#63220) Homepage Journal

  #!/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?

  Reply to This

  • Re: (Score:2)

    by Ovid (2709)

    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
    • Re: (Score:1)

      by dakkar (5832)

      Argh! *three numbers*, not *two pairs*. Stupid typing-faster-than-reading...
  • Re: (Score:1)

    by rjbs (4671)

    First, you need a "grep { $_ !~ /0/ }" in there. Second, I think that your three numbers, concatenated, must use each of the digits only once all together. That is, a valid answer would be qw(123 456 789) if only those numbers divided properly. qw(123 331 882) is no good.

    --
    rjbs
    • Re: (Score:2)

      by Ovid (2709)

      That appears to be correct. You have to read the spec carefully. They ask you to use each of those numbers, not any. I got that wrong the first time.
      • Re: (Score:2)

        by AndyArmstrong (7200)

        Christ on a bike - I'm rubbish at reading specifications :)