What is the remainder when (31^31)^101 divided by 9?

• (31^3131)/9
  =((27+4)^3131)/9
  =(4^3131)/9
  =(((4^3)^1043)*4^2)/9
  =((64^1043)*16)/9
  =((63+1)^1043)*16)/9
  =((1^1043)*16)/9
  =16/9
  7 remainder ans.
  
  shukriya kehne ki zarurat nhi hai.....bs duwaon mein yaad rkhna...
• 11 years agoHelpfull: Yes(34) No(3)
• 31*101=3131
  (27+4)^3131/9
  4^3131/9
  2^6262/9
  (2^3)^2087*2/9
  8^2087*2/9
  (9-1)^2087*2/9
  -1*2/9
  -2/9
  ans:9-2=7
• 11 years agoHelpfull: Yes(6) No(2)
• 31/9 gives remainder 4
  4^31^101/9=(((4^(3*10)*4)^101/9=64/9 gives remainder 1
  1^101=1 and again 4^101/9=(4^(3*33))*4^2/9=16/9=7 remainder
• 11 years agoHelpfull: Yes(4) No(0)
• 4 would be the answer
• 11 years agoHelpfull: Yes(2) No(2)
• 31*101=3131
  (31)/9=R(4)
  thus (4)^3131=4
  because (4)^odd=4.
  hence ans=4
• 11 years agoHelpfull: Yes(1) No(5)
• #ans: 7
  (using common division rules)
• 11 years agoHelpfull: Yes(1) No(1)
• (31^31)101=31^3131
  taking smallest ramainder dividing 31/9= 31^3131/9=4^3131/9
  find out cyclicity
  ist term 4/9=rem=4
  second term 4^2/9=16/9=7
  3rd term 4^3/9=7*4/9=1
  since cylicity of 3
  then 3131/3=2 term of cycleis reaminder=7
  
• 11 years agoHelpfull: Yes(1) No(1)
• 31^7 devided by 3 so 31^14 devided by 9,so ans is ZERO
  
• 11 years agoHelpfull: Yes(0) No(3)
• (((31)^31)^101)/9
  =(((9*3+4)^31)^101)/9
  =(((4)^31)^101)/9
  =(((4^3)^10* 4)^101)/9
  =(((64)^10*4)^101)/9
  =(((9*7+1)^10*4)^101)/9
  =((1^10*4)^101)/9
  =((4^3)^33*4^2)/9
  =((9*7+1)^33*16)/9
  =((1^33)*16)/9= 16/9
  Remainder = 7
• 11 years agoHelpfull: Yes(0) No(1)