fermat's number

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fermat's number
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.

 Posts: 125
 Joined: Mon Dec 13, 2010 12:05 pm
 Location: চট্রগ্রাম,Chittagong
 Contact:
Re: fermat's number
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..........
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..........
Re: fermat's number
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=$2^{2^5}+1$
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.
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.
Re: fermat's number
This is due to Euler.
Note that $641=5^4+2^42^{32}+5^42^{28}$ and $641=5.2^7+15^42^{28}1$
So $641$ divides their difference $2^{32}+1$
Note that $641=5^4+2^42^{32}+5^42^{28}$ and $641=5.2^7+15^42^{28}1$
So $641$ divides their difference $2^{32}+1$
One one thing is neutral in the universe, that is $0$.
Re: fermat's number
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.
so ,you can see that 4294967297/641...........and that is your ans.
Re: fermat's number
i dont think you will have access to wikipedia or calculator in any MATH contest .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.
Re: fermat's number
But this is not allowed in the olympiad and you need to prove this rigorouslyDipan 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.
And we need the proof,you just checked this.
One one thing is neutral in the universe, that is $0$.