jmc

We know that $6^4>999_{10}>6^3$. So, we can tell that $999_{10}$ in base six will have four digits. $6^3=216$, which can go into 999 four times at most, leaving $999-4\cdot 216=135$ for the next three digits. $6^2=36$ goes into 135 three times at most, leaving us with $135-3\cdot 36=27$. Then, $6^1=6$ goes into 27 four times at most, leaving $27-4\cdot 6=3$ for the ones digit. All together, the base six equivalent of $999_{10}$ is $\boxed{4343_6}$.