Mathematics · Binomial Theorem

JEE Main 2025 — 8 April, Evening Shift — Question 42

The product of the last two digits of $(1919)^{1919}$ is $\_\_\_\_$ -

Answer: 63

Numerical answer — enter this value.

$(1919)^{1919}=(1920-1)^{1919}$

\begin{array}{r} ={ }^{1919} C_{0}(1920)^{1919}-{ }^{1919} C_{1}(1920)^{1918}+\ldots+ { }^{1919} C_{1918}(1920)-{ }^{1919} C_{1919} \end{array}

For last two digit $\Rightarrow{ }^{1919} C_{1919}(1920)-1$ $=3684479$

$\therefore \quad$ Product of last two digit $=63$

Answer key and solution verified before publishing.

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