functions
\(\large \color{black}{\begin{align} & f(x)=x^4+x^3+x^2+x+1,\ \ x\in \mathbb{Z}^{>1} \hspace{.33em}\\~\\ & \normalsize \text{Find remainder when }\ \large f(x^5)\ \normalsize \text{is divided by} \ \large f(x) ? \hspace{.33em}\\~\\ & a.)\ 1 \hspace{.33em}\\~\\ & b.)\ 4 \hspace{.33em}\\~\\ & c.)\ 5 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{a monomial in }\ x \hspace{.33em}\\~\\ & e.)\ \normalsize \text{a polynomial in }\ x \hspace{.33em}\\~\\ \end{align}}\)
Rewrite \(f(x) = \dfrac{x^5-1}{x-1}\) and \(f(x^5) = \dfrac{x^{25}-1}{x^5 - 1}\)
Don't we need \(|x|<1\) for that to be true? @ParthKohli
Nope, not at all.
Oh right I'm thinking of infinite sums.
Notice that \(f(x) = \dfrac{x^5-1}{x-1} \implies x^5-1=f(x)*(x-1)\tag{1}\) \[\begin{align}f(x)&=x^4+x^3+x^2+x+1 \\~\\ \implies f(x^5)&=x^{20}+x^{15}+x^{10}+x^5+1\\~\\ &=(x^{20}-1)+(x^{15}-1)+(x^{10}-1)+(x^5-1)+5\\~\\ &=(x^5-1)(stuff)+5\\~\\ &=f(x)*(x-1)*(stuff)+5 ~~~\color{gray}{\text{(from (1))}}\\~\\ &\equiv 0+5\pmod{f(x)} \end{align}\]
wow thnx!
look up problem #30
oh u remember all things fro past cool.
*from
looks very handy book
i will solve it within 10 years. haha
try these https://drive.google.com/folderview?id=0B8qeUE5SqcPAWFVaM1N5anN3S2M&usp=sharing
its showing "error 500"
it has a very good collection of olympiad textbooks
yes got it .
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