fermat's number

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AntiviruShahriar
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fermat's number

Unread post by AntiviruShahriar » Tue Dec 14, 2010 2:16 pm

Prove that 5th Fermat's number $2^{2^5}+1$ is a multiple of 641.
Last edited by AntiviruShahriar on Wed Dec 15, 2010 1:28 pm, edited 2 times in total.

AntiviruShahriar
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Re: fermat's number

Unread post by AntiviruShahriar » Tue Dec 14, 2010 2:44 pm

i tried to write 2^2^5 as dollar($){2^(2^5)+1}dollar($) but didn't worked!!!!!!!!!!someone help me to write like that......
about problem:i used modular and after about restless 8 days i made the solution of it.........now after yours solution I'll post my one............it was the hardest problem I'd ever solved..........

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Moon
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Re: fermat's number

Unread post by Moon » Tue Dec 14, 2010 3:13 pm

You have towrite 2^{2^5}+1 (I think you made mistake last time) ;)
I mean when you want to write something bigger than 1 letter you have to use second bracket.
So, 2^{2^5}+1=$2^{2^5}+1$ :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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Masum
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Re: fermat's number

Unread post by Masum » Wed Dec 15, 2010 11:56 am

This is due to Euler.
Note that $641=5^4+2^4|2^{32}+5^42^{28}$ and $641=5.2^7+1|5^42^{28}-1$
So $641$ divides their difference $2^{32}+1$
One one thing is neutral in the universe, that is $0$.

Dipan
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Re: fermat's number

Unread post by Dipan » Wed Dec 29, 2010 12:19 pm

Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.

tushar7
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Re: fermat's number

Unread post by tushar7 » Wed Dec 29, 2010 1:34 pm

Dipan wrote:Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.
i dont think you will have access to wikipedia or calculator in any MATH contest .

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Masum
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Re: fermat's number

Unread post by Masum » Wed Dec 29, 2010 1:47 pm

Dipan wrote:Ok, this is an easy problem..if you read in wikipedia about FERMAT NUMBER you can get the answer........2^2^5=2^32=4294967297
so ,you can see that 4294967297/641...........and that is your ans.
But this is not allowed in the olympiad and you need to prove this rigorously 8-)
And we need the proof,you just checked this.
One one thing is neutral in the universe, that is $0$.

Dipan
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Re: fermat's number

Unread post by Dipan » Wed Dec 29, 2010 10:31 pm

Sorry.. :cry:

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