\(\frac{321_4}{23_4}\\=\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\\=\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\\=\frac{48+8+1}{8+3}\\=\frac{57}{11}=5\hspace{1mm}remainder\hspace{1mm}2\\∴r=2_{10} \\ Now\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\\\frac{4}{2} = 2\\\frac{4}{0}=0\hspace{1mm}or\hspace{1mm}2\\∴r=2\)