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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Binomial Theorem

The product of the last two digits of (1919)^ 1919 is

The product of the last two digits of (1919)1919(1919)^{1919} is \qquad

Correct answer

63

Step-by-step solution

(1919)1919=(1920−1)1919=1919C0(1920)1919−1919C1(1920)1918+….+1919C1918(1920)1−1919C1919=100λ+1919×1920−1=100λ+3684480−1=100λ+………..79 (last two digit) ⇒ Number having last two digit 79∴ Product of last two digit 63\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 \\ & =100 \lambda+\ldots \ldots \ldots . .79 \text{ (last two digit) } \\ & \Rightarrow \text{ Number having last two digit } 79 \\ & \therefore \text{ Product of last two digit } 63\end{aligned}
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