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(D) Let $P(x) = 1+x^2+x^4+\ldots+x^{2010}$.This is a geometric progression with $1006$ terms,first term $a=1$,and common ratio $r=x^2$.$P(x) = \frac{1

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$1006$

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(D) Let $P(x) = 1+x^2+x^4+\ldots+x^{2010}$.This is a geometric progression with $1006$ terms,first term $a=1$,and common ratio $r=x^2$.$P(x) = \frac{1-(x^2)^{1006}}{1-x^2} = \frac{1-x^{2012}}{1-x^2}$.We want $Q(x) = 1+x+x^2+\ldots+x^{n-1} = \frac{1-x^n}{1-x}$ to divide $P(x)$.$P(x) = \frac{(1-x^{1006})(1+x^{1006})}{(1-x)(1+x)} = \left(\frac{1-x^{1006}}{1-x}ight) \left(\frac{1+x^{1006}}{1+x}ight)$.Note that $\frac{1-x^{1006}}{1-x} = 1+x+x^2+\ldots+x^{1005}$.For $Q(x)$ to divide $P(x)$,$n$ must be a divisor of $1006$ such that $1005 \le n \le 2010$.The divisors of $1006 = 2 	imes 503$ are $1, 2, 503, 1006$.The only divisor in the interval $[1005, 2010]$ is $n=1006$.However,checking the condition $Q(x) | P(x)$,we require $n$ to be such that the roots of $Q(x)$ are roots of $P(x)$.The roots of $Q(x)$ are $e^{i \frac{2k\pi}{n}}$ for $k=1, 2, \ldots, n-1$.For these to be roots of $P(x) = \frac{1-x^{2012}}{1-x^2}$,we need $x^{2012}=1$ but $x^2 
eq 1$.This implies $n$ must divide $2012$ and $n$ must not divide $2$.Given the interval $[1005, 2010]$,the only value satisfying the divisibility is $n=1006$ (if we consider the structure of the polynomial division). Re-evaluating,the correct answer is $n=1006$.

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 $1+x^2+x^4+x^6+\ldots+x^{2010}$ is

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