Let the number $(22)^{2022}+(2022)^{22}$ leave the remainder $\alpha$ when divided by 3 and $\beta$ when divided by 7. Then $\left(\alpha^{2}+\beta^{2}\right)$ is equal to :

We have, $(22)^{2022}+(2022)^{22}$ <br/><br/>As 2022 is completely divisible by 3 <br/><br/>So, $(2022)^{22}$ is also divisible by 3 <br/><br/>$(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$ <br/><br/>$\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 . <br/><br/>$\therefore(22)^{2022}+(2022)^{22}$ leave a remainder when divisible by 3 <br/><br/>$\therefore \alpha=1$ <br/><br/>$$ \begin{aligned} (22)^{2022}+(2022)^{22} & =(21+1)^{2022}+(2023-1)^{22} \\\\ & =7 K+1+7 \mu+1=7(K+\mu)+2 \end{aligned} $$ <br/><br/>$\Rightarrow(22)^{2022}+(2022)^{22}$ leave a remainder 2 when divisible by 7 <br/><br/>$$ \begin{aligned} & \therefore \beta=2 \\\\ & \text { Hence, } \alpha^2+\beta^2=1^2+2^2=5 \end{aligned} $$