The remainder of $$\frac{{{{32}^{{{32}^{32}}}}}}{7}:$$

Solution (By Examveda Team)

$$\eqalign{ & {32^{{{32}^{32}}}}\,{\text{means}}\,{32^{\left( {32.32.32.32.......32\,\text{times}} \right)}} \cr & {\text{and}}\,{\text{the}}\,{\text{remainder}}\,{\text{of}}\,\frac{{32}}{7}\,{\text{is}}\,4 \cr & {\text{So}}, \cr & \frac{{{4^{\left( {32.32.32.32.......32\,\text{times}} \right)}}}}{7} \cr & \frac{{{4^{\left( {2.2.2.2.2........32\,\text{times}} \right)}}}}{7} \cr & {\text{Remainder}} = 4 \cr & {\text{Since}},\,\frac{4}{7} \to \,{\text{Remainder}}\,4 \cr & \frac{{{4^2}}}{7} \to \,{\text{Remainder}}\,2 \cr & \frac{{{4^3}}}{7} \to \,{\text{Remainder}}\,1 \cr & \frac{{{4^4}}}{7} \to \,{\text{Remainder}}\,4 \cr} $$