JEE Mains · Maths · STD 11 - 7. binomial theoram
The remainder when \((2021)^{2022}+(2022)^{ 2021 }\) is divided by \(7\) is.
- A \(0\)
- B \(1\)
- C \(2\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\((2021)^{\text {a022 }}+(2022)^{2011}\) \(=(2023-2)^{\text {an2 }}+(2023-1)^{3031}\) \(=7 n_{1}+2^{m 2 n 2}+7 n _{2}-1\) \(=7\left( n _{1}+ n _{2}\right)+8^{674}-1\) \(=7\left(n_{1}+n_{2}\right)+(7-1)^{674}-1\) \(=7\left(n_{1}+n_{2}\right)+7 n_{3}+1-1\)…
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