find remainder when 2^89 is divided by 89

• 2^89/89 = (2^88)*(2^1)/89
  = (2^11)^8 * (2^1)/89
  = (2048^8) * (2^1)/89
  = (89*23+1)^8 * (2^1) / 89
  = 1^8 * 2^1 / 89
  = 2 /89
  = 2
• 10 years agoHelpfull: Yes(24) No(4)
• power of 2 has 4 cycles
  so 2^89====>89/4==22(remainder is 1)
  which implies 2^89===some number that ends with 2
  2^89==>_______2
  when this is divided by 89 ans may be 0 or 2
• 10 years agoHelpfull: Yes(7) No(3)
• when we take successive powers of 2 and find their remainders, we get the following cyclic patterns of cycle length 11.
  viz 2,4,8,16,32,64,39,78,67,45,1
  i.e 2^11 leaves a remainder 1.
  
  Thus 289=(2^11)^8*(2) leaves a remainder of 2.
• 10 years agoHelpfull: Yes(3) No(1)
• remainder of 2^89 when divided by 89
  =rem(rem(2^88)*rem(2))
  =rem((rem(2^11)*rem(2^11)...8 times)*rem(2))
  =rem(rem(1*1*...8 times)*rem(2))
  =rem(1*2)
  =2
• 10 years agoHelpfull: Yes(1) No(3)
• cyclicity of 2 is =4then 2^89%4=2^1,then 2^1/89 =reminder 2
• 8 years agoHelpfull: Yes(1) No(0)
• 39
  128/89=remaindr 39
  
• 10 years agoHelpfull: Yes(0) No(19)