Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremRemainder and Divisibility ProblemsMedium2 minPYQ_2022

MathematicsMediumsingle choice

The remainder when ${(2021)}^{2022} + {(2022)}^{2021}$ is divided by $7$ is

Options:

Answer:

Solution:

We know $2023 = 7 \times 289$
So ${(2021)}^{2022} + {(2022)}^{2021}$
$= {(2023 - 2)}^{2022} + {(2023 - 1)}^{2021}$
$= 7 m_{1} + 2^{2022} + 7 m_{2} + {(- 1)}^{2021}$
$= 7 (m_{1} + m_{2}) + 8^{674} - 1$
$= 7 (m_{1} + m_{2}) + (7 + 1)^{674} - 1$
$= 7 (m_{1} + m_{2}) + 7 m_{3} + 1 - 1$
$= 7 (m_{1} + m_{2} + m_{3})$
This is a multiple of $7$
Hence, remainder is $0$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Remainder and Divisibility Problems

⏱ 2mℹ️ Source: PYQ_2022

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