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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jul 21, 2006 9:06 am Post subject:
Re: Question about exponention



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(2 = phi 4 * phi 7 = 2 * 6 = 12
5, 28 coprime
5^12 = 1 (mod 28)
5^49 = 5^(4*12) * 5 = 5 (mod 2
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

Posted: Fri Jul 21, 2006 8:25 am Post subject:
Re: Question about exponention



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

Posted: Fri Jul 21, 2006 8:13 am Post subject:
Question about exponention



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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