Let the number ( 22 ) 2022 + ( 2022 ) 22 leave the remainder when divided by 3 and when divided by 7. Then ( 2 + 2 ) is equal to [2023]

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Solution

Q.
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 [2023]

1 20

2 13

3 5

4 10

Ans.
(3)

We are, ${(22)}^{2022} + {(2022)}^{22} = {(21 + 1)}^{2022} + {(2022)}^{22}$

Here ${(2022)}^{22}$ is divisible by 3 as 2022 is divisible by 3.

Expanding ${(21 + 1)}^{2022}$ , we get

${(21 + 1)}^{2022} = {}^{2022}{C_{0}} {(21)}^{2022} + {}^{2022}{C_{1}} {(21)}^{2021} + \dots + {}^{2022}{C_{2022}} {(1)}^{2022}$

   $= [{(3)}^{2022} {(7)}^{2022} + {}^{2022}{C_{1}} {(3)}^{2021} 7^{2021} + \dots] + 1$

   $= 3 [{(3)}^{2021} {(7)}^{2022} + {}^{2022}{C_{1}} {(3)}^{2020} {(7)}^{2021} + \dots] + 1$

   $= 3 m + 1 ∴ Remainder, α = 1$

Also,   ${(21 + 1)}^{22} + {(2023 - 1)}^{22}$

$= 7 [{(3)}^{2022} {(7)}^{2021} + {}^{2022}{C_{1}} {(3)}^{2021} {(7)}^{2020} + \dots] + 1 + {}^{22}{C_{0}} {(2023)}^{22} - {}^{22}{C_{1}} {(2023)}^{21} + \dots - {}^{22}C_{21} (2023) + {}^{22}{C_{22}}$

   $= (7 r_{1} + 1) + 7 [(7^{21}) {(289)}^{22} - {}^{22}{C_{1}} {(7)}^{20} {(289)}^{21} + \dots - {}^{22}{C_{21}} (289)] + 1$

$= (7 r_{1} + 1) + (7 r_{2} + 1) = 7 (r_{1} + r_{2}) + 2 = 7 n + 2$

$∴ Remainder, β = 2$

Now, $α^{2} + β^{2} = 1^{2} + 2^{2} = 5$

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