remainder of 3^1000 divided by 73

• 3^1 / 73 results in remainder of 3
  3^2 / 73 results in remainder of 9
  3^3 / 73 results in remainder of 27
  3^4 / 73 results in remainder of 8
  
  continuing same way we get a sequence of 3,9,27,8,24..... at 3^13/73 we have remainder equals to 3
  i.e the sequence repeats at cycle of 12
  
  hence 1000/12 gives a remainder 4
  remainder 4 here implies to the 4th term of the sequence which is 8.
  hence 8 is the required solution
  
  
• 7 years agoHelpfull: Yes(10) No(0)
• ans =8
  3^1000/73=
  =(3^4)250/73
  =81^250/73
  =8^250/73(remainder theorem)
  =(8^3)^83 *8/73
  =512^83/73
  =1^83*8/73
  =8 ans
  
• 7 years agoHelpfull: Yes(7) No(8)
• (3^4)^250/73=(81^250)/73=(8^250)/73=((8^3)^83*8)/73=8
• 7 years agoHelpfull: Yes(2) No(8)
• ans=8
  3^1000/73=
  from "fermet theorem" 3^72/73=1remainder; { x^(n-1)/n=1(remainder)}
  so 3^(72*13)*3^64/73 3^72*13/73=1rem
  = 3^64/73
  = (3^4)*16/73 81/73=8rem
  = 8^16/73
  = (8^4)*4/73 8^4/73=8rem
  = 8^4/73
  gives remainder=8 ans
• 7 years agoHelpfull: Yes(2) No(1)
• 29 is the answer
  3^1000/73=
  =(3^4)250/73
  =(81*250)/73
  =((73+8)*250)/73
  =250+(2000)/73
  =250+(1460+540)/73
  =250+20+540/73
  =270+(73*7+29)/73
  =277+29/73
  so 29 is the answer
• 7 years agoHelpfull: Yes(1) No(9)
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• 7 years agoHelpfull: Yes(0) No(4)
• 26 1000^3 and 1000 divided by 73 gives same remainder. So,1000 divided by 73 gives 26
• 7 years agoHelpfull: Yes(0) No(1)