{(29^31)^109}/9

• step 1st :
  take (31^109)=N
  step 2nd:
  solve (29^N)/9
  =on which 2 get reminder on dividing 29/9 so,
  (2^N)/9
  step 3rd:
  now find the power cycle when remainder became 1 as
  (2^1)/9=2(remainder)
  (2^2)/9=4(remainder)
  (2^3)/9=8/9= -1 or 8
  so remainder -1 get on 3rd cycle so 1 get in 6th cycle:
  step 4:
  using 6 as power cycle we divide N
  p== N/6=(31^109)/6=(1^109)/6 as 31/6=1 as per reminder theorem
  step 5:
  now put p in step 2 as value of N..
  {(29^31)^109}/9==(2^p)/9
  =(2^1)/9 as p=1
  =2 as per reminder theorem
  so final remainder is 2..
• 7 years agoHelpfull: Yes(7) No(0)
• (29^31) => 9^1(odd number)=9
  (29^31)^109 => 9^9(odd number)=9
  (29^31)^109/9 => 9/9=1;
  thus the reminder becomes zero
• 7 years agoHelpfull: Yes(2) No(7)
• First step 29/9 =reminder is 2 geting so {(2^31)^109}/9
  Second step =>31*109= 3379 => (2^3379)/9
  Third step=> 2^6/9 gives the reminder 1 so divide 3379/6 we will get reminder as 1(Here 6 is as 2^1/9=2, 2^2/9=4 like power cycle is 6)
• 7 years agoHelpfull: Yes(0) No(15)