The remainder of $$\frac{{{{32}^{{{32}^{32}}}}}}{7}:$$
A. 1
B. 2
C. 3
D. 4
E. None of these
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {32^{{{32}^{32}}}}\,{\text{means}}\,{32^{\left( {32.32.32.32.......32\,\text{times}} \right)}} \cr & {\text{and}}\,{\text{the}}\,{\text{remainder}}\,{\text{of}}\,\frac{{32}}{7}\,{\text{is}}\,4 \cr & {\text{So}}, \cr & \frac{{{4^{\left( {32.32.32.32.......32\,\text{times}} \right)}}}}{7} \cr & \frac{{{4^{\left( {2.2.2.2.2........32\,\text{times}} \right)}}}}{7} \cr & {\text{Remainder}} = 4 \cr & {\text{Since}},\,\frac{4}{7} \to \,{\text{Remainder}}\,4 \cr & \frac{{{4^2}}}{7} \to \,{\text{Remainder}}\,2 \cr & \frac{{{4^3}}}{7} \to \,{\text{Remainder}}\,1 \cr & \frac{{{4^4}}}{7} \to \,{\text{Remainder}}\,4 \cr} $$Join The Discussion
Comments ( 1 )
Related Questions on Number System
Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
A. 12, 24, 36
B. 11, 22, 33
C. 12, 24, 32
D. 5, 10, 15
4^2/7= 16/7 rem = 2
4^3/7= 64/7, rem = 1
Sol. Have gone nuts 😶
Correct me if im wrong.