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Given $1^2+2^2+3^2+....+2003^2$ = $(2003)(4007)(334)$ and $(1)(2003)+(2)(2002)+(3)(2001)+....+(2003)(1)$ = $(2003)(334)(x)$ then value of $x$ is

$(a)\;2001\qquad(b)\;2003\qquad(c)\;2005\qquad(d)\;2004$

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