Q28 / 75 · JEE Main 2025 (8 Apr Shift 2) · Mathematics
The product of the last two digits of $(1919)^{1919}$ is
$\qquad$
Correct answer: 63
Solution
$\begin{aligned} & (1919)^{1919}=(1920-1)^{1919} \\ & ={ }^{1919} \mathrm{C}_0(1920)^{1919}-{ }^{1919} \mathrm{C}_1(1920)^{1918}+\ldots . \\ & +{ }^{1919} \mathrm{C}_{1918}(1920)^1-{ }^{1919} \mathrm{C}_{1919} \\ & =100 \lambda+1919 \times 1920-1 \\ & =100 \lambda+3684480-1 \\ & =100 \lambda+\ldots \ldots \ldots . .79 \text { (last two digit) } \\ & \Rightarrow \text { Number having last two digit } 79 \\ & \therefore \text { Product of last two digit } 63\end{aligned}$
Hence, the correct answer is 63.