(31^31)^728/9.. what is the remainder?

• (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
• 10 years agoHelpfull: Yes(20) No(2)
• ((31)^31)^728/9
  = ((9*3+4)^31)^728)/9=((4)^31)^728/9
  =((4)^3*4)^728/9=((64)^10*4)^728/9=((9*7+1)^10*4)^728/9=(4)^728/9
  =((4)^3)^242*4^2/9 = 4^2/9
  =16/9 = 7(remainder)
• 10 years agoHelpfull: Yes(13) No(8)
• 31*31*728 =
  4 , 4, -1
  -16.
  ans is 7
  short process
  
• 10 years agoHelpfull: Yes(7) No(2)
• answer will be 4
  (31^31)^728/9
  so 31/9= reminder 4
  so (4^31)^728/9
  so (4*4^30)^728/9= 4*(4^2)^15^728/9=4*(16^15)^728/9
  so 16/9= reminder 7
  4*(7^15)^728/9= 4*7*(7^14)^728/9=4*7(4^7)^728/9=4*7*4*(4^6)^728/9=4*7*4(7^3)^728/9=4*7*4*7(4^728)/9= reminder 4
• 10 years agoHelpfull: Yes(6) No(13)
• answer-7.
  explanation-->(31^31)^728/9 can be written as (4^31)^728/9
  now 4 when divided by 9 has got a cycle of 3.now 728x31=22568 which on dividing by 3 gives 2 as remainder which means that the second term in the cycle i.e 4^2 mod 9=7 is the answer.
  
• 10 years agoHelpfull: Yes(6) No(0)
• By using power cycles concept 31^11 will lead to 31 in last 2 digits thus we can convert the given expression to (31^11*31^11*31^9)^729/9 thus the last 2 digits will be (31*31*31^9)^729/9=31^729/9=(31^11*66)*31^2/9=31^3/9=29791/9 thus the remainder is 1
• 10 years agoHelpfull: Yes(4) No(7)
• (31^31)^728=31^22868/9= 4^22868/9=7
• 10 years agoHelpfull: Yes(3) No(2)
• 31^31=961^728=(960+1)^728=1^728/9=1
• 10 years agoHelpfull: Yes(2) No(12)
• (31^1)^2/9=7 is the remainder
• 10 years agoHelpfull: Yes(2) No(0)
• ans will be 4
  ((27+4)^31)^728/9
  =(4^31)^728
  
  (4^31)%9=4
  
  (4^728)%9=4
  
  so ans 4
• 10 years agoHelpfull: Yes(1) No(8)
• answr is 7
• 10 years agoHelpfull: Yes(1) No(1)
• answer will be 7.
• 10 years agoHelpfull: Yes(0) No(5)
• (31^31)^728 /9 rem=1
  31^4/9=29791/9 rem=1
  
  also rem (31^4)^31*182=(1)^(31*182)= rem =1
  
  ans rem=1
• 10 years agoHelpfull: Yes(0) No(1)
• 31=27+4
  (27+4)^31^728/9
  (4)^31^728/9
  31%8=7
  (4)^7^728
  728%4=0
  (4)7^0
  4^1/9
  ans=4
• 10 years agoHelpfull: Yes(0) No(0)