Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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Unread post by faisalnir » Tue Feb 01, 2011 1:51 pm

$$F(x)=x^{n}+x^{n-1}+.................+x^2+x+2$$ find the remainder when $$F(x^{n+1})$$ is divided by $$F(x)$$.

Mehfuj Zahir
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Re: remainder

Unread post by Mehfuj Zahir » Wed Feb 02, 2011 3:29 pm

u can try with geometric progression

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Re: remainder

Unread post by Cryptic.shohag » Sun Feb 06, 2011 1:54 pm

faisalnir wrote:$$F(x)=x^{n}+x^{n-1}+.................+x^2+x+2$$ find the remainder when $$F(x^{n+1})$$ is divided by $$F(x)$$.
\[\therefore f(x^{n+1})=(x^{n+1})^n+(x^{n+1})^{n-1}+........+(x^{n+1})^2+x^{n+1}+2\]

Now, \[f(x)=\, \frac{x(x^n-1)}{x-1}+2\]\[=\, {P}x(x-1)+2\]

And, \[f(x^{n+1})=\, \frac{x^{n+1}((x^{n+1})^n-1)}{x^{n+1}-1}+2\]\[=\, {P}'x^{n+1}(x^{n+1}-1)+2\]
If we observe it's clear that {P}' is a multiple of P

Now, \[f(x^{n+1})-f(x)=\, {P}'x^{n+1}(x^{n+1}-1)+2-({P}x(x-1)+2)=\, KPx^{n+1}{P}''(x-1)-{P}x(x-1)=Px(x-1)(K{P}''x^{n+1}-1)\]

If we divide \[Px(x-1) \: by \: f(x), \: the \: remainder\: will \: be \: -2\]. So, if we divide \[(K{P}''x^{n+1}-1) by\: f(x)\] the remainder will be \[-2\times (K{P}''x^{n+1}-1)\]. But I can't find the value of \[(K{P}''x^{n+1}-1)\]. All I can guess is that the remainder could be \[-2\]....... :( :( :(
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein

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