Find the remainder when $3^{1999}$ is divided by $13$.
Since $3^3=27=2\cdot13+1$ we find that  \[3^3\equiv1\pmod{13}.\] Therefore \[3^{1999}\equiv3^{3\cdot666+1}\equiv1^{666}\cdot3\equiv3\pmod{13}.\] The remainder when $3^{1999}$ is divided by 13 is $\boxed{3}$.