jmc

Since $3^3=27=2\cdot 13+1$ we find that $$3^3\equiv 1(\mathrm{mod}13).$$ Therefore $$3^{1999}\equiv 3^{3\cdot 666+1}\equiv 1^{666}\cdot 3\equiv 3(\mathrm{mod}13).$$ The remainder when $3^{1999}$ is divided by 13 is $\boxed{3}$.