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Q. The sum of last two digits of the number $7^{101}$ is ___

Binomial Theorem

Solution:

$7^{101} =7(49)^{50}=7(1-50)^{50} $
$=7\left[{ }^{50} C_{0}-{ }^{50} C_{1} 50+{ }^{50} C_{2} 50^{2}-\ldots\right] $
$=7[1+100 \lambda] $ where $\lambda \in N=7+700 \lambda$
Therefore, last two digits of number $7^{101}$ are $07$ .