331. Find the last two digits of the number $137^{42}$.
331. The problem reduces to finding the remainder of the division of the number $137^{42}$ by 100. Since 137 and 100 are coprime, $137^{\varphi(100)}-1$ is divisible by $100, \quad$ but $\varphi(100)=$ $=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40 ; \quad$ therefore, $\quad 137^{40}-1$ is divisible by 100. Hence, $137^{42}=137^{2}\left(137^{40}-1\right)+137^{2}=$ $=137^{2} \cdot\left(137^{40}-1\right)+(100+37)^{2}=100 n+37^{2}=100 n+$ $+1369=100 n+1300+69$, from which it is immediately clear that the desired remainder is 69.