Let $S$ be the [set](https://artofproblemsolving.com/wiki/index.php/Set) of the $2005$ smallest positive multiples of $4$, and let $T$ be the set of the $2005$ smallest positive multiples of $6$. How many elements are common to $S$ and $T$?
$\textbf{(A) } 166\qquad \textbf{(B) } 333\qquad \textbf{(C) } 500\qquad \textbf{(D) } 668\qquad \textbf{(E) } 1001$
Since the [least common multiple](https://artofproblemsolving.com/wiki/index.php/Least_common_multiple) $\mathrm{lcm}(4,6)=12$, the [elements](https://artofproblemsolving.com/wiki/index.php/Element) that are common to $S$ and $T$ must be [multiples](https://artofproblemsolving.com/wiki/index.php/Multiple) of $12$. 
Since $4\cdot2005=8020$ and $6\cdot2005=12030$, several multiples of $12$ that are in $T$ won't be in $S$, but all multiples of $12$ that are in $S$ will be in $T$. So we just need to find the number of multiples of $12$ that are in $S$. 
Since $4\cdot3=12$, every $3$rd element of $S$ will be a multiple of $12$.
Therefore the answer is $\left \lfloor\frac{2005}{3} \right \rfloor=\boxed{\textbf{(D) }  668}$