34. Find the last 2 digits of
$$
x=19^{17^{15^{3}}}
$$
34. Answer: 59
Let $\alpha=17^{15}$ and $\beta=15^{133^{3^{1}}}$. Since $13 \quad$ is odd. $\beta \equiv-1(\bmod 16)$. Now let $\varphi$ be Euler's phi function, $\varphi(100)=40$ and $\varphi(40)=16$. By Euler's theorem,
$$
\alpha=17^{9} \equiv 17^{-1} \equiv 33 \quad(\bmod 40)
$$
where the last congruence can be calculated by the extended Euclidean algorithm. Thus by repeated squaring, we have
$$
19^{\alpha} \equiv 19^{33} \equiv 59 \quad(\bmod 100)
$$