JEE Main 2025 Apr 8 Shift 2, Mathematics Q24: Introduction to Binomial Expansion

JEE Main2025Apr 8, Shift 2Mathematics

The product of the last two digits of $(1919)^{1919}$ is ______

Your answer

Solution

Write $1919 = 1920 - 1$ and expand $(1920 - 1)^{1919}$ using the binomial theorem.

Every term except the last two contains a factor of at least $192 0^{2}$ , which is divisible by 100, contributing nothing to the last two digits.

The last two terms are $(1918 1919) (1920) (- 1)^{1918} + (1919 1919) (- 1)^{1919} = 1919 \cdot 1920 - 1$ .

Compute: $1919 \times 1920 = 3684480$ , so the sum contributed is $3684480 - 1 = 3684479$ .

The full number is $100 k + 3684479$ for some integer $k$ , so the last two digits come from $3684479 mod 100 = 79$ .

Product of these digits: $7 \times 9 = 63$ .

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