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

$7^{101}=7(49{)}^{50}=7(1-50{)}^{50}$

$\begin{array}{l}=7\left[{ }^{50}{C}_0{-}^{50}{C}_{1}50{+}^{50}{C}_2{50}^2-\dots \right]\\ =7[1+100\lambda ]\text{ where }\lambda \in N=7+700\lambda \end{array}$

$\text{ Therefore, last two digits of number }{7}^{101}\text{ are }07\text{ . }$