2^214^302 when divided by 9 find remainder? *mark q

• when dealing with large power try to express in terms of the divisor say here for exmaple:
  2^214^302=64^208^302.
  now 64= 9*7 + 1
  so inner expression wud give a remainder as 1,which when raised to any power wud be 1. so the anwer is 1. if i am wrng please rectify .
• 11 years agoHelpfull: Yes(30) No(18)
• 2^6=64 when 64/9 then remainder is 1
  2^(214*302)
  when 214*302 divided by 6 then ramainder will be 2
  1*2^2=4
  hence remaider will be 4..
• 11 years agoHelpfull: Yes(25) No(25)
• Powers of 2 when divided by 9 gives remainder in form a series of 5 terms
  2 4 8 7 5
  Now 2^214 is of form 2^5n+4 so it will give remainder 7 when divided by 9.
  2^214 * 2^214 * 2^214 * 2^214............upto 302 terms
  7 * 7 *..................................upto 302 terms
  
  Now when powers of 7 divided by 9 they forms a series of 3 terms 4 1 7
  So 7^302 = 7^ (3n + 2)
  Hence when it is divided by 9 the remainder would be 1.
  
• 11 years agoHelpfull: Yes(11) No(3)
• 7
  we can write 214^302 in the form (3k+1).
  Now 2^(3k+1)...where k is always odd and we can write {2^(3k)*2^1}
  when we find the remainder it comes to 7
  
  
• 11 years agoHelpfull: Yes(9) No(14)
• 7 is the answer
  
• 11 years agoHelpfull: Yes(6) No(3)
• 1 is the answer
• 11 years agoHelpfull: Yes(4) No(7)
• answer is 7
• 11 years agoHelpfull: Yes(2) No(0)
• 7
  2^1/9 gives reminder = 7
  2^2/9 gives reminder = 5
  2^3/9 gives reminder = 1
  2^4/9 gives reminder = 7
  2^5/9 gives reminder = 5
  2^6/9 gives reminder = 1
  hence reminder repeats periodically...
  214^302/3 gives reminder=1
  hence ans=7
• 11 years agoHelpfull: Yes(1) No(3)
• Euler no of 9 is 6.
  (9=3^2
  Euler no 9=9*23 =6)
  so divide 214^302 by 6, which will remainder 1.
  therefore 2^1 gives 2, which on dividing by 9 gives 2.
  
• 11 years agoHelpfull: Yes(1) No(2)
• How to solve these kind of problems,plz nyone kindly xplain in detail...
• 11 years agoHelpfull: Yes(0) No(7)
• can anyone explain in detail plz?
• 11 years agoHelpfull: Yes(0) No(0)
• ans: 1
  214*302=64628
  64628%4=0
  so 2^0=1
• 11 years agoHelpfull: Yes(0) No(1)
• ={(2^6)^35*2^2}^302/9
  ={(64)^35*2^2}^302/9
  =[1*4]^302/9
  =[(4^3)^100*4^2]/9
  =16/9
  =7(Rem)
• 9 years agoHelpfull: Yes(0) No(0)
• by remainder theorem answer is 1
  
• 9 years agoHelpfull: Yes(0) No(0)
• remainders are
  2^1/9=2
  2^2/9= 4
  2^3/9= 8
  2^4/9= 7
  2^5/9= 5
  2^6/9 = 1
  2^7/9=2
  cyclicity of 6 times therefore 214/6=4 is the reminder and in cyclicity fourth remainder term is 7. so 7 is the ans is this helpful
• 9 years agoHelpfull: Yes(0) No(0)
• Using the simple rule of {(a^x)/(a+1)}
  when x is odd, remainder is a
  when x is even, remainder is 1
  So using this rule and solving we get remainder as 1
• 9 years agoHelpfull: Yes(0) No(0)