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Mathematics 30 Online
OpenStudy (mathmath333):

functions

OpenStudy (mathmath333):

\(\large \color{black}{\begin{align} & f(x)=x^4+x^3+x^2+x+1,\ \ x\in \mathbb{Z}^{>1} \hspace{.33em}\\~\\ & \normalsize \text{Find remainder when }\ \large f(x^5)\ \normalsize \text{is divided by} \ \large f(x) ? \hspace{.33em}\\~\\ & a.)\ 1 \hspace{.33em}\\~\\ & b.)\ 4 \hspace{.33em}\\~\\ & c.)\ 5 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{a monomial in }\ x \hspace{.33em}\\~\\ & e.)\ \normalsize \text{a polynomial in }\ x \hspace{.33em}\\~\\ \end{align}}\)

Parth (parthkohli):

Rewrite \(f(x) = \dfrac{x^5-1}{x-1}\) and \(f(x^5) = \dfrac{x^{25}-1}{x^5 - 1}\)

OpenStudy (anonymous):

Don't we need \(|x|<1\) for that to be true? @ParthKohli

Parth (parthkohli):

Nope, not at all.

OpenStudy (anonymous):

Oh right I'm thinking of infinite sums.

ganeshie8 (ganeshie8):

Notice that \(f(x) = \dfrac{x^5-1}{x-1} \implies x^5-1=f(x)*(x-1)\tag{1}\) \[\begin{align}f(x)&=x^4+x^3+x^2+x+1 \\~\\ \implies f(x^5)&=x^{20}+x^{15}+x^{10}+x^5+1\\~\\ &=(x^{20}-1)+(x^{15}-1)+(x^{10}-1)+(x^5-1)+5\\~\\ &=(x^5-1)(stuff)+5\\~\\ &=f(x)*(x-1)*(stuff)+5 ~~~\color{gray}{\text{(from (1))}}\\~\\ &\equiv 0+5\pmod{f(x)} \end{align}\]

OpenStudy (mathmath333):

wow thnx!

ganeshie8 (ganeshie8):

look up problem #30

OpenStudy (mathmath333):

oh u remember all things fro past cool.

OpenStudy (mathmath333):

*from

OpenStudy (mathmath333):

looks very handy book

OpenStudy (mathmath333):

i will solve it within 10 years. haha

OpenStudy (mathmath333):

its showing "error 500"

ganeshie8 (ganeshie8):

it has a very good collection of olympiad textbooks

OpenStudy (mathmath333):

yes got it .

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