When 2³¹ is divided by 5 remainder is -

$${{\text{2}}^{31}} \div 5$$
\[\begin{gathered} {\text{power }}\,\,\,\,\,\,\,\,\,{\text{ remainder}} \hfill \\ \left[ \begin{gathered} {2^1}\,\,\, \to \,\,\,\frac{2}{5} \to \,\,\,2 \hfill \\ {2^2}\,\,\, \to \,\,\,\frac{4}{5} \to \,\,\,4 \hfill \\ {2^3}\,\,\, \to \,\,\,\frac{8}{5} \to \,\,\,3 \hfill \\ {2^4}\,\,\, \to \,\,\,\frac{{16}}{5} \to \,\,1 \hfill \\ \end{gathered} \right]{\text{cycle 1}} \hfill \\ \left[ {{2^5}\,\,\, \to \,\,\,\frac{{243}}{5} \to \,\,2} \right]{\text{cycle 2}} \hfill \\ \end{gathered} \]

$$\eqalign{ & {\text{Divided = }}\frac{{{2^{31}}}}{5} = \frac{{{2^{4 \times 7}} \times {2^3}}}{5} \cr & \Rightarrow {\text{remainder = 3}} \cr & {\text{So }}{{\text{2}}^3}{\text{ has 3 remainder }} \cr & {{\text{2}}^{31}}{\text{ has 3 remainder}} \cr} $$