Answer: Option B
$${{\text{2}}^{31}} \div 5$$
\[\begin{gathered}
{\text{power }}\,\,\,\,\,\,\,\,\,{\text{ remainder}} \hfill \\
\left[ \begin{gathered}
{2^1}\,\,\, \to \,\,\,\frac{2}{5} \to \,\,\,2 \hfill \\
{2^2}\,\,\, \to \,\,\,\frac{4}{5} \to \,\,\,4 \hfill \\
{2^3}\,\,\, \to \,\,\,\frac{8}{5} \to \,\,\,3 \hfill \\
{2^4}\,\,\, \to \,\,\,\frac{{16}}{5} \to \,\,1 \hfill \\
\end{gathered} \right]{\text{cycle 1}} \hfill \\
\left[ {{2^5}\,\,\, \to \,\,\,\frac{{243}}{5} \to \,\,2} \right]{\text{cycle 2}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Divided = }}\frac{{{2^{31}}}}{5} = \frac{{{2^{4 \times 7}} \times {2^3}}}{5} \cr
& \Rightarrow {\text{remainder = 3}} \cr
& {\text{So }}{{\text{2}}^3}{\text{ has 3 remainder }} \cr
& {{\text{2}}^{31}}{\text{ has 3 remainder}} \cr} $$