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Question about exponention
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William Elliot
science forum Guru


Joined: 24 Mar 2005
Posts: 1906

PostPosted: Fri Jul 21, 2006 9:06 am    Post subject: Re: Question about exponention Reply with quote

On Fri, 21 Jul 2006 WingDragon@gmail.com wrote:

Quote:
sorry, I have missed something...
should be:
2^(5^49) mod 29 = ?

Golly, another ##$@%^ computer problem.


2^28 = 1 (mod 29)

phi(2Cool = phi 4 * phi 7 = 2 * 6 = 12
5, 28 coprime
5^12 = 1 (mod 28)

5^49 = 5^(4*12) * 5 = 5 (mod 2Cool
2^(5^49) = 2^5 = 32 = 3 (mod 29)

Quote:
WingDragon@gmail.com 寫道:

How can I calculate such a large value like this?
2^(5^49) = ?

I can't calculate the result by any calculator.

That's because calculators are stupid.
Can they calculate the simple equivalence?
25 = 5 (mod 5)

Quote:
Is there any way to find the answer more easily?

Yes, but only if you dare to turn off the computer and do your own
thinking.
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WingDragon@gmail.com
science forum beginner


Joined: 21 Jul 2006
Posts: 2

PostPosted: Fri Jul 21, 2006 8:25 am    Post subject: Re: Question about exponention Reply with quote

sorry, I have missed something...
should be:
2^(5^49) mod 29 = ?


WingDragon@gmail.com 寫道:

Quote:
How can I calculate such a large value like this?
2^(5^49) = ?

I can't calculate the result by any calculator. Is there any way to
find the answer more easily?
Thanks and sorry for my poor English.

Fred
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WingDragon@gmail.com
science forum beginner


Joined: 21 Jul 2006
Posts: 2

PostPosted: Fri Jul 21, 2006 8:13 am    Post subject: Question about exponention Reply with quote

How can I calculate such a large value like this?
2^(5^49) = ?

I can't calculate the result by any calculator. Is there any way to
find the answer more easily?
Thanks and sorry for my poor English.

Fred
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Google

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