The remainder when $(11{)}^{1011}+(1011{)}^{11}$ is divided by 9 is

$$\begin{gathered} (11{)}^{1011}+(1011{{)}^{11}={\left(9+2\right)}^{11}+\left(1008+3\right)}^{11} \\ =9k+2^{1011+3^{11}}=9k+8^{337}+3^{11} \\ =9k+{\left(9-1\right)}^{337}+3{\left(9\right)}^5 \\ =9k-1==9k-9+8=9\left(Integer\right)+8 \\ \mathrm{Remainder} \mathrm{when} (11{)}^{1011}+(1011{)}^{11} \mathrm{is} \mathrm{divided} \mathrm{by} 9 \mathrm{is} 8 \end{gathered}$$