jmc

We know that $4^4>123_{10}>4^3$. So, we can tell that $123_{10}$ in base four will have four digits. $4^3=64$, which can go into 123 only one time at most, leaving $123-1\cdot 64=59$ for the next three digits. $4^2=16$ goes into 59 three times at most, leaving us with $59-3\cdot 16=11$. Then, $4^1=4$ goes into 11 two times at most, leaving $11-2\cdot 4=3$ for the ones digit. All together, the base four equivalent of $123_{10}$ is $\boxed{1323_4}$.