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The number of natural numbers $n$ in the interval $[1005,2010]$ for which the polynomial $1+x+x^{2}+x^{3}+\ldots . x^{n-1}$ divides the polynomial $1+x^{2}+x^{3}+x^{4}+\ldots .+x^{2010}$ is-
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Verified Answer
The correct answer is:
503
$$
\begin{array}{l}
1+x^{2}+x^{4}+\ldots \ldots x^{2010}=\frac{1\left(1-x^{2012}\right)}{1-x^{2}}=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)} \\
=\left(1+x^{1006}\right)\left(\frac{\left(1-x^{503}\right)}{(1-x)}\right)\left(\frac{\left(1+x^{503}\right)}{(1+x)}\right) \\
=\left(1+x^{1006}\right)\left(1+x+x^{2}+\ldots . x^{502}\right)\left(1-x+x^{2}-x^{3}+\ldots . x^{502}\right)
\end{array}
$$
this is divisible by $1+x+x^{2}+\ldots . . x^{n-1}$
if $n-1=502$
$$
n=503
$$
\begin{array}{l}
1+x^{2}+x^{4}+\ldots \ldots x^{2010}=\frac{1\left(1-x^{2012}\right)}{1-x^{2}}=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)} \\
=\left(1+x^{1006}\right)\left(\frac{\left(1-x^{503}\right)}{(1-x)}\right)\left(\frac{\left(1+x^{503}\right)}{(1+x)}\right) \\
=\left(1+x^{1006}\right)\left(1+x+x^{2}+\ldots . x^{502}\right)\left(1-x+x^{2}-x^{3}+\ldots . x^{502}\right)
\end{array}
$$
this is divisible by $1+x+x^{2}+\ldots . . x^{n-1}$
if $n-1=502$
$$
n=503
$$
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