On Sun, May 4, 2008 at 12:36 PM, Marcelo <marcelo.magallon@[...].com>  wrote:

>  On Sun, May 4, 2008 at 6:22 AM, Harry Kakueki <list.push@[...].com> wrote:
> 
>  >  ruby q161.rb 10000000000000000
>  >  84 numbers
> 
>  I've got 106 in that range:
> 
>  $ time ./reverse_divisible 10_000_000_000_000_000
>  106
> 
>  real    0m1.457s
>  user    0m1.364s
>  sys     0m0.092s
> 
>  The only reason why I don't post the solution yet is because I'm stuck
>  on a proof that the method is correct :-(
> 
>  Marcelo
> 
> 
Here's my attempt:
class Integer
  def divisible_by_reverse?
    return false unless self % 9 == 0
    return false if self % 10 == 0
    reverse = self.to_s.reverse
    return false if self.to_s.eql?(reverse)
    return self % reverse.to_i == 0
  end
end

max = ARGV[0] ? ARGV[0].to_i : 1_000_000
1.upto(max) { |n| puts n if n.divisible_by_reverse? }

-- 
Shane Emmons
E: semmons99@gmail.com