JEE Mains · Maths · STD 11 - 7. binomial theoram
The remainder when \(19^{200}+23^{200}\) is divided by \(49\) , is \(.........\).
- A \(28\)
- B \(27\)
- C \(29\)
- D \(26\)
Answer & Solution
Correct Answer
(C) \(29\)
Step-by-step Solution
Detailed explanation
\((21+2)^{200}+(21-2)^{200}\) \(\Rightarrow 2\left[{ }^{100} C _0 21^{200}+200 C _2 21^{198} \cdot 2^2+\ldots . .+{ }^{200} C _{198} 21^2\right.\) \(\left.2^{198}+2^{200}\right]\) \(\Rightarrow 2\left[49 I _1+2^{200}\right]=49 I _1+2^{201}\) Now,…
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