JEE Mains · Maths · STD 11 - 7. binomial theoram
The product of the last two digits of \((1919)^{1919}\) is ___________.
- A 60
- B 75
- C 63
- D 70
Answer & Solution
Correct Answer
(C) 63
Step-by-step Solution
Detailed explanation
\begin{aligned} & (1919)^{1919}=(1920-1)^{1919} \\ & ={ }^{1919} \mathrm{C}_0(1920)^{1919}-{ }^{1919} \mathrm{C}_1(1920)^{1918}+\ldots . \\ & +{ }^{1919} \mathrm{C}_{1918}(1920)^1-{ }^{1919} \mathrm{C}_{1919} \\ & =100 \lambda+1919 \times 1920-1 \\ & =100 \lambda+3684480-1 \\ &…
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