2002 AMC 12B Problem 21

Since $2002=11\cdot13\cdot14$ with $11,$ $13,$ $14$ pairwise coprime, $a_n=11$ when $n$ is divisible by $182,$ $a_n=13$ when $n$ is divisible by $154,$ and $a_n=14$ when $n$ is divisible by $143$ (and no $n\lt2002$ is divisible by all three). For $n\le2001$ there are $10$ multiples of $182,$ $12$ of $154,$ and $13$ of $143.$ So the sum is $$\begin{gathered} 11\cdot10+13\cdot12+14\cdot13 \\ {}=110+156+182 \\ {}=448. \end{gathered}$$ Thus, the correct answer is A.