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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

  1. A \(10\)
  2. B \(5\)
  3. C \(20\)
  4. D \(13\)
Verified Solution

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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