(32^32^32)/7
what will be the remainder

• remainder=4
  
  (32^32^32)/7
  using remainder theorem---
  
  (((4)^32)^32)/7.............32/7 gives rem 4
  ((((4)^2)^16)^32)/7
  (((16)^16)^32)/7.............16/7 gives rem 2
  (((2)^16)^32)/7
  ((2(2)^15)^32)/7
  ((2(8)^5)^32)/7
  ((2)^32 x 1)/7............ 8/7 gives rem 1
  (4(8)^10)/7
  4 x 1 ................ 8/7 gives rem 1
  =4 !!!!!!
  
• 9 years agoHelpfull: Yes(0) No(1)
• ANS WILL BE 1.
  DIVING 32 BY 7,WE GET 4 AS REMAINDER.
  NOW THE SITUATION IS:(4*4*4)/7=(16*4)/7
  AGAIN DIVING 16 BY 7,2 IS THE REMAINDER
  (2*4)/7=8/7 WHICH GIVES 1 AS REMAINDER.
• 9 years agoHelpfull: Yes(0) No(3)
• (32^32^32)/7 using remainder theorem
  32/7 = 4 remainder
  now (4^32^32)/7= (4^2^16^32)/7 =(2^16^32)/7=(2^4^32)/7=(2^32)= gives remainder 2^2 = 4
• 9 years agoHelpfull: Yes(0) No(0)