Question

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

Answer 1

Answer: 7

$7^{101}=7(49{)}^{50}=7(1-50{)}^{50}$
$=7\left[{}^{50}{C}_0-{}^{50}{C}_{1}50+{}^{50}{C}_{2}50^2-\dots \right]$
$=7[1+100\lambda ]$ where $\lambda \in N=7+700\lambda$
Therefore, last two digits of number $7^{101}$ are $07$ .