use Perl

• Second try (Score:1)

  by colomon (8994)

  I like this version better: http://paste.lisp.org/display/84391 [lisp.org]
  • Re: (Score:1)

    by colomon (8994)

    But it's a bit slower for some reason.
• care to contribute to the euler_bench program? (Score:1)

  by notbenh (7967)

  {after some poking from Eric Wilhelm I'll post this here as well}

  I'm part of PDX.pm and we've started a project to collect solutions to euler problems as a way to benchmark rakudo. We would love to have you join in the fun. Currently everything is up on github (http://github.com/notbenh/euler_bench/tree/master).

  --
  benh~
  • Re: (Score:1)

    by colomon (8994)

    I will add what I've got in as soon as I can puzzle out git. (A good exercise for me anyway!)
• The answer has to be divisible by 9 (Score:1)

  by btilly (5037)

  The sum of the digits of a number mod 9 is always the number mod 9. The sum of the digits of 2*$n and 3*$n is the same, which means that mod 9, 2*$n and 3*$n are the same, which means that $n is 0 mod 9, so $n is divisible by 9.

  This should improve your speed by a factor of 3. :-)
  • Re: (Score:1)

    by colomon (8994)

    Huh. I don't follow your logic -- I don't know anything about sum of the digits being the same means the numbers are the same mod 9. BUT a web search trying to turn up more on it found a proof that "The difference of any two numbers composed of the same digits is always a multiple of nine." In which case 3*$n - 2*$n = $n is a multiple of nine, so your conclusion is dead on correct. Thanks!