fermat's number

For students of class 9-10 (age 14-16)
AntiviruShahriar
Posts:125
Joined:Mon Dec 13, 2010 12:05 pm
Location:চট্রগ্রাম,Chittagong
Contact:
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
Posts:125
Joined:Mon Dec 13, 2010 12:05 pm
Location:চট্রগ্রাম,Chittagong
Contact:

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..........

User avatar
Moon
Site Admin
Posts:751
Joined:Tue Nov 02, 2010 7:52 pm
Location:Dhaka, Bangladesh
Contact:

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.

User avatar
Masum
Posts:592
Joined:Tue Dec 07, 2010 1:12 pm
Location:Dhaka,Bangladesh

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
Posts:158
Joined:Wed Dec 08, 2010 5:36 pm

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
Posts:101
Joined:Tue Dec 07, 2010 3:23 pm

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 .

User avatar
Masum
Posts:592
Joined:Tue Dec 07, 2010 1:12 pm
Location:Dhaka,Bangladesh

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
Posts:158
Joined:Wed Dec 08, 2010 5:36 pm

Re: fermat's number

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

Sorry.. :cry:

Post Reply