find the remainder of 32^32^32 when divided by 9

• 32 = 2⁵
  (2⁵)³² = 2¹⁶⁰
  (2¹⁶⁰)³² = 2⁵¹²⁰
  You are looking for the remainder of 2⁵¹²⁰ ÷ 9.
  
  2¹ mod 9 = 2
  2² mod 9 = 4
  2³ mod 9 = 8
  2⁴ mod 9 = 7
  2⁵ mod 9 = 5
  2⁶ mod 9 = 1
  2⁷ mod 9 = 2
  2⁸ mod 9 = 4
  2⁹ mod 9 = 8
  2¹⁰ mod 9 = 7
  2¹¹ mod 9 = 5
  2¹² mod 9 = 1
  ...
  Note that the pattern repeats, and when the exponent is divisible by 6, the remainder is 1.
  5120 = 6(853) + 2, therefore
  2⁵¹²⁰ mod 9 = 4
• 10 years agoHelpfull: Yes(3) No(2)
• The answer is 7
• 10 years agoHelpfull: Yes(1) No(0)
• K VEDANT LASKAR
  PLEASE POST THE SOLUTION
• 9 years agoHelpfull: Yes(0) No(0)
• 32^32^32/9 can be written as
  5^32^32
  25^16^32
  7^16^32
  49^8^32
  4^8^32
  16^4^32
  7^4^32
  49^2^32
  4^2^32
  16^32
  7^32
  49^16
  4^16
  16^8
  7^8
  49^4
  4^4
  16^2
  7^2
  49when divided by 9gives 4 as a ans
• 8 years agoHelpfull: Yes(0) No(0)