JEE Mains · Maths · STD 11 - 7. binomial theoram
Remainder when \( 64^{32^{32}}\) is divided by \(9\) is equal to ...........
- A \(5\)
- B \(4\)
- C \(8\)
- D \(1\)
Answer & Solution
Correct Answer
(D) \(1\)
Step-by-step Solution
Detailed explanation
Let \(32^{32}=\mathrm{t}\) \( 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 \mathrm{t}} \) \( =9 \mathrm{k}+1\) Hence remainder \(=1\)
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