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  4. The last four digits of the natural number 7100 are

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Q. The last four digits of the natural number $7^{100}$ are

Binomial Theorem

Solution:

$7^{100} = 49^{50} = \left(50 - 1\right)^{50} = \left(1 - 10\right)^{50}$
$= 1-50\times50+\frac{50\cdot49}{2}\cdot\left(50\right)^{2}-\frac{50\cdot49\cdot48}{6}\cdot\left(50\right)^{3}$
+ multiple of $10^{4}$
$= 1 - 2500 + 3062500 +$ multiple of $10^{4}$
$= 3060001 +$ multiple of $10^{4}$
$\therefore $ The last four digits are $0001$.