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## All the Perl that's Practical to Extract and Report

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• #### Second try(Score:1)

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

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

{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)

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)

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)

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!