5. Find the last two digits (in order) of $7^{30105}$.
5. Ans: 43
Note that $7^{4}=1 \quad(\bmod 100)$. Now
$$
3^{2005}=(-1)^{2005}=-1=3 \quad(\bmod 4) .
$$

Hence $3^{2005}=4 k+3$ for some positive integer $k$. Thus
$$
7^{3^{2005}}=7^{4 k+3}=7^{3}=43 \quad(\bmod 100)
$$