32^32^32/9 leaves the remainder?
i)2
ii)4
iii)1
iv)7

• 4 will be the answer. because 32^32^32/9 = 2^160^32/9 = 7^32/9 = 4
• 11 years agoHelpfull: Yes(4) No(1)
• 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
• 11 years agoHelpfull: Yes(2) No(0)
• 32^32^32/9=(27+5)^32^32/9
  rem=5^32^32/9=(27-2)^16^32
  rem=2^16^32=[2(9-1)^3]^32/9
  rem=2^32/9=4(9-1)^10
  so finally reminder=4...........
• 11 years agoHelpfull: Yes(1) No(0)