JEE Mains · Maths · STD 11 - 7. binomial theoram
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
- A \(10\)
- B \(5\)
- C \(20\)
- D \(13\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\((22)^{2022}+(2022)^{22}\) divided by \(3\) \((21+1)^{2022}+(2022)^{22}\) \(=3 k+1 \quad(\alpha=1)\) Divided by \(7\) \((21+1)^{2022}+(2023-1)^{22}\) \(7 k +1+1 \quad(\beta=2)\) \(7 k +2\) So \(\alpha^2+\beta^2 \Rightarrow 5\)
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