The product of the last two digits of $(1919)^{1919}$ is
Answer (integer)
63
Solution
<p>$$\begin{aligned}
& (1919)^{1919}=(1920-1)^{1919} \\
& ={ }^{1919} \mathrm{C}_0(1920)^{1919}-{ }^{1919} \mathrm{C}_1(1920)^{1918}+\ldots . \\
& +{ }^{1919} \mathrm{C}_{1918}(1920)^1-{ }^{1919} \mathrm{C}_{1919} \\
& =100 \lambda+1919 \times 1920-1 \\
& =100 \lambda+3684480-1 \\
& =100 \lambda+\ldots \ldots \ldots . .79 \text { (last two digit) } \\
& \Rightarrow \text { Number having last two digit } 79 \\
& \therefore \text { Product of last two digit } 63
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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