20. Find the last two digits of $2011^{\left(2010^{2000)}\right.}$.
20. Ans: 01 .
Note that $2011 \equiv 11(\bmod 100)$ and $11^{2} \equiv 21,11^{3} \equiv 31(\bmod 100)$ etc. So $11^{10} \equiv 1$ (mod 100). Since $2010^{2009}$ is divisible by 10 ,
$$
2011^{2010^{200}} \equiv 11^{10} \times \cdots \times 11^{10} \equiv 1 \quad(\bmod 100) .
$$