8. Determine the last two digits of $7^{5^{6}}$.
(A) 01
(B) 07
(C) 09
(D) 43
(E) 49
8. Answer: (B)
Note that $7^{4}-1=2400$, so that $7^{4 n}-1$ is divisible by 100 for any $n \in \mathbb{Z}^{+}$. Now,
$$
\begin{aligned}
7^{5^{6}} & =7\left(7^{5^{5}-1}-1+1\right) \\
& =7\left(7^{5^{5}-1}-1\right)+7 \\
& =7\left(7^{4 n}-1\right)+7
\end{aligned}
$$
where
$$
n=\frac{5^{6}-1}{4} \in \mathbb{Z}^{+}
$$

Since $7\left(7^{4 n}-1\right)$ is divisible by 100 , its last two digits are 00 . It follows that the last two digits of $7^{5^{6}}$ are 07 .