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20379Re: [PrimeNumbers] Too big for any computer?

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  • Norman Luhn
    Jun 9 7:45 AM
      I agree with Phil.

      Calculate 3^53,106 -> 3
      then 2^3+99 = 107 , 107 =0 mod 107


      --- Phil Carmody <thefatphil@...> schrieb am Di, 9.6.2009:

      Von: Phil Carmody <thefatphil@...>
      Betreff: Re: [PrimeNumbers] Too big for any computer?
      An: "Prime Number" <primenumbers@yahoogroups.com>
      Datum: Dienstag, 9. Juni 2009, 16:37

      --- On Tue, 6/9/09, Devaraj Kandadai <dkandadai@gmail. com> wrote:

      > I had  requested Ken (Kradenken)  to test whether  2^(3^53) + 99   is

      > eactly divisible by  107 or not.  He replied that

      > the number is too big for any computer.  Do you agree?

      If by "the number" you mean 2^(3^53)+99, then clearly it isn't too big for any computer as it can trivially be represented using only 9 characters. It's binary or decimal expansion is too big, but no-one's asking for its binary or decimal expansion, so that's irrelevant.

      To test whether it's divisible by 107 is trivial also, as 2^phi(107) == 1 (mod 107). So reduce 3^53 modulo phi(107):

      ? Mod(2,107)^lift( Mod(3,eulerphi( 107))^53) +99

      Mod(0, 107)

      So indeed, it is divisible by 107.


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