32^32^32 if divided by 7.what will be the remainder?

• At first,one point should be cleared that 32^32^32 and (32^32)^32 carry different meanings from each other.And according to the prob. we have to work with 32^32^32.
  Now,32 =(7*4)+4
  So,{(7*4)+4}^32^32 = (4^32^32)mod7.
  Now, observe that 32^32 =(33-1)^32 which is definitely of the form (3A+1),where A is a +ve integer,but A's value isn't important to know here.
  SO,
  (4^32^32)mod7 = (4^{3A+1})mod7 ={[4^(3A)]*4}mod7 ={[64^A]*4}mod7 =
  {[(7*9 +1)^A]*4}mod7 = [(1^A)*4]mod7 = 4mod7 .
  
  So,the remainder will be 4 .
• 12 years agoHelpfull: Yes(11) No(4)
• 32^32^32 if divided by 7
  first take 32^32 means 32 is a number having power 32
  32^32 mod 7=(7*4+4)^32=(7*4)^32 will give remainder 0,so take only (4)^32
  ((4)^2)^16 mod 7=(16)^16=(7*2+2)^16 mod 7=(2)^16 mod 7=((2)^4)^4 mod 7=(7*2+2)^4 mod 7=(2)^4 mod 7=8 mod 7=1
  now next
  (1)^32=1 will be remainder(ans)
• 12 years agoHelpfull: Yes(4) No(9)