QJEEM23Ma1711: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1711 Solution: Let the number ( 22 ) 2022 + ( 2022 ) 22 leave the remainder α…

Question

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Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $α$ when divided by 3 and $β$ when divided by 7. Then $(α^{2} + β^{2})$ 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 μ + 1 = 7 (K + μ) + 2 \end{aligned}

\Rightarrow(22)^{2022}+(2022)^{22}

leave a remainder 2 when divisible by 7

\begin{aligned} ∴ β = 2 \\ Hence, α^{2} + β^{2} = 1^{2} + 2^{2} = 5 \end{aligned}

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