jmc

It is much easier to find $120_4÷2_4$ and then multiply by $13_4$ than it is to do the calculations in the original order. For $120_4÷2_4$, we have $$\begin{array}{cccc}\text{multicolumn}2r& 3& 0& \\ \text{cline}2-42& 1& 2& 0\\ \text{multicolumn}2r1& 2& \downarrow & \\ \text{cline}2-3\text{multicolumn}2r& 0& 0& \\ \text{multicolumn}2r& 0& 0& \\ \text{cline}3-4\text{multicolumn}2r& & 0& \end{array}$$for a quotient of $30_4$. Note that $12_4÷2_4=6_{10}÷2_{10}=3_{10}=3_4$. Now we find the product of $13_4$ and $30_4$. \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & \stackrel{2}{1} & 3_4& \\ & & & \times & & 3 & 0_4 \\ \cline{4-7} & & &1 &1 & 1 & 0_4 \\ \end{array}The answer is $\boxed{1110_4}$.