BdMO National Higher Secondary 2007/7

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

BdMO National Higher Secondary 2007/7

Unread post by BdMO » Sun Feb 06, 2011 10:18 pm

Problem 7:
$f(x)=x^6+x^5+\cdots +x+1$Find the remainder when dividing $f(x^7)$ by $f(x)$.

checkmatec4
Posts: 2
Joined: Tue Feb 08, 2011 12:53 pm

Re: BdMO National Higher Secondary 2007/7

Unread post by checkmatec4 » Tue Feb 08, 2011 5:23 pm

I am quite sure that the solution is the following:
f(x)= 1+x+x^2+x^3+.....+x^6
Now if we write it like this:
f(x)= 1+x+x^2+x^3+.....infinite, then
f(x)= (1-x)^(-1)
f(x^7)=(1-x^7)^(-1)
so f(x^7)/f(x) = (1-x)/(1-x^7)
according to remainder theorem, the remainder is = 0
so Answer: 0.
(Osman, a friend of mine solved this)

Ehsan
Posts: 6
Joined: Tue Jan 03, 2012 6:17 pm

Re: BdMO National Higher Secondary 2007/7

Unread post by Ehsan » Tue Feb 07, 2012 5:03 pm

Actually that is applicable when x is less than 1 but this was not given in the question, and I'm looking for a solution of this one long time so I think some of the seniors could help

User avatar
nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

Re: BdMO National Higher Secondary 2007/7

Unread post by nafistiham » Tue Feb 07, 2012 9:18 pm

Hint:
\[f(x)\cdot(x-1)=x^7-1\]
solution
\[f(x^7)=x^{42}+x^{35}+x^{28}+x^{21}+x^{14}+x^7+1\]
\[\Rightarrow f(x^7)=(x^{42}-1)+(x^{35}-1)+(x^{28}-1)+(x^{21}-1)+(x^{14}-1)+(x^7-1)+7\]
here, $x^7-1=f(x)\cdot(x-1)$
so, \[f(x)|(x^{42}-1),(x^{35}-1),(x^{28}-1),(x^{21}-1),(x^{14}-1),(x^7-1)\]
so, the remainder will be
\[7\]
;)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

Pinku71
Posts: 2
Joined: Fri Jun 03, 2011 10:01 pm

Re: BdMO National Higher Secondary 2007/7

Unread post by Pinku71 » Fri Feb 10, 2012 12:39 am

nice and clean:D

sakib.creza
Posts: 26
Joined: Sat Nov 03, 2012 6:36 am

Re: BdMO National Higher Secondary 2007/7

Unread post by sakib.creza » Thu Feb 07, 2013 9:03 pm

nafistiham wrote: here, $x^7-1=f(x)\cdot(x-1)$
so, \[f(x)|(x^{42}-1),(x^{35}-1),(x^{28}-1),(x^{21}-1),(x^{14}-1),(x^7-1)\]
so, the remainder will be
\[7\]
How does the green line lead to the red line? :? :? :?: :?:

User avatar
nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

Re: BdMO National Higher Secondary 2007/7

Unread post by nafistiham » Thu Feb 07, 2013 10:45 pm

sakib.creza wrote:
nafistiham wrote: here, $x^7-1=f(x)\cdot(x-1)$
so, \[f(x)|(x^{42}-1),(x^{35}-1),(x^{28}-1),(x^{21}-1),(x^{14}-1),(x^7-1)\]
so, the remainder will be
\[7\]
How does the green line lead to the red line? :? :? :?: :?:
actually, what I didn't think much important to mention was this
\[f(x)|x^{7}-1|x^{7n}-1\]
It should be clear enough now
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

Post Reply