I agree with Phil.

Calculate 3^53,106 -> 3

then 2^3+99 = 107 , 107 =0 mod 107

best

--- 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.

Phil

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