## BdMO National Higher Secondary 2007/7

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National Higher Secondary 2007/7

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

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
(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

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

nafistiham
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### Re: BdMO National Higher Secondary 2007/7

Hint:
solution
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

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

### Re: BdMO National Higher Secondary 2007/7

nice and clean:D

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

### Re: BdMO National Higher Secondary 2007/7

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?

nafistiham
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### Re: BdMO National Higher Secondary 2007/7

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?
$f(x)|x^{7}-1|x^{7n}-1$
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.