(31^31)^728/9 remainder?

• (31^31)^728/9
  = (4^31)^728/9
  =(64^10*4)^728/9
  =4^728/9
  =64^242*16/9
  =16/9
  =7
• 11 years agoHelpfull: Yes(22) No(6)
• rem of ((31^31)^728)/9= 1
  as 31^4= 29791 this when divided by 9 leaves 1 remainder.
  as 728 is divisible by 4 therefore rem of ((31^31)^728)/9= 1
  ans=1
• 11 years agoHelpfull: Yes(4) No(7)
• (((31)^31)^728)/9
  =(((9*3+4)^31)^728)/9
  =(((4)^31)^728)/9
  =(((4^3)^10* 4)^728)/9
  =(((64)^10*4)^728)/9
  =(((9*7+1)^10*4)^728)/9
  =((1^10*4)^728)/9
  =((4^3)^242*4^2)/9
  =((9*7+1)^242*16)/9
  =((1^242)*16)/9
  Remainder = 7
• 11 years agoHelpfull: Yes(4) No(0)
• 4. 31/9 gives 4.so ((4^3)^10)^728 /9 gives remainder 1 and rest 4^4.182 gives 4.4*1=4
• 11 years agoHelpfull: Yes(2) No(6)
• ans is 7 not 1
• 11 years agoHelpfull: Yes(2) No(0)
• (27+4)^31^728
  since 27 is divisible completely by 9,we calculate remainder when 4^31^728 is divided by 9
  4%9=4
  4^2%9=7
  4^3%9=1
  4^4%9=4
  and thus cycle repeats at 4,7,1
  31*728=22568
  22568%3=2
  so ans would be 4^2 % 9=7
• 11 years agoHelpfull: Yes(2) No(0)
• sorry the remainder will be 7.
  
• 11 years agoHelpfull: Yes(1) No(2)
• 31*728 = 22568
  now 31^22568
  (27+4)^22568
  => 4^22568
  2^45136
  8^15042
  (9-1)^15042
  -1/9
  = 8 ans
• 11 years agoHelpfull: Yes(0) No(5)
• the remainder will be 9
• 11 years agoHelpfull: Yes(0) No(3)
• according to euler's theorem the asn is: 4
• 11 years agoHelpfull: Yes(0) No(1)
• ((3*9 +4)^31)^728
  =(4^31)^728
  =((4^4)^27)^728
  =((63+1)^27)^728
  =(1^27)^728
  =1
  
• 11 years agoHelpfull: Yes(0) No(1)
• (31^31)^728/9=(31^31)^728/3*3=(1^31)^728/3=1/3=1. so ans is 1.
• 11 years agoHelpfull: Yes(0) No(1)
• --------correction in pervious ans---------
  rem of ((31^31)^728)/9= 1
  as 31^3= 29791 this when divided by 9 leaves 1 remainder.
  as 728 is divisible by 3
  (31^30)^728 * (31)^728= rem(1)* rem(1)
  ans 1
• 11 years agoHelpfull: Yes(0) No(1)