Let the number (22)^2022+(2022)^22 leave the remainder when divided by 3 and when divided …

Question

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

Accepted Answer

We have, $(22)^{2022}+(2022)^{22}$

As 2022 is completely divisible by 3

So, $(2022)^{22}$ is also divisible by 3

$(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$

$\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 .

$\therefore(22)^{2022}+(2022)^{22}$ leave a remainder when divisible by 3

$\therefore \alpha=1$

$$
\begin{aligned}
(22)^{2022}+(2022)^{22} & =(21+1)^{2022}+(2023-1)^{22} \\
& =7 K+1+7 \mu+1=7(K+\mu)+2
\end{aligned}
$$

$\Rightarrow(22)^{2022}+(2022)^{22}$ leave a remainder 2 when divisible by 7

$$
\begin{aligned}
& \therefore  \beta=2 \\
& \text { Hence, }  \alpha^2+\beta^2=1^2+2^2=5
\end{aligned}
$$

Solutions · 1