JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 3)
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 :
13
10
20
5
Explanation
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
$$ \begin{aligned} & \therefore \beta=2 \\\\ & \text { Hence, } \alpha^2+\beta^2=1^2+2^2=5 \end{aligned} $$
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
$$ \begin{aligned} & \therefore \beta=2 \\\\ & \text { Hence, } \alpha^2+\beta^2=1^2+2^2=5 \end{aligned} $$
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