what is the remainder when 30^72^87 is divided by 11.

• 30^72^87 mod 11
  E[11] = 10
  Then
  72^87 mod 10
  2^87 mod 10
  2^1 mod 10 = 2
  2^2 mod 10 = 4
  2^3 mod 10 = 8
  2^4 mod 10 = 6
  2^5 mod 10 = 2 again...
  Cyclicity is 4... R[2^87/10] ll be 4k+n th value...
  87 => 4*21 + 3
  2^87 mod 10 = 8
  Now the expression is reduced to 30^8 mod 11
  8^8 mod 11
  2^24 mod 11
  E[11] = 10
  (2^10)^2 * 2^4 mod 11
  1 * 16 mod 11
  5 mod 11
  
  Ans : 5
• 9 years agoHelpfull: Yes(21) No(7)
• ANSWER will be definitely 7
  30^72^87 mod 11
  =(-3)^72^87 mod 11
  =(9)^36^87 mod 11
  =(-2)^36^87 mod 11
  =16^9^87 mod 11
  =5^9^87 mod 11
  =(5^3*5^3*5^3)^87 mod 11
  =(4*4*4)^87 mod 11
  =(64)^87 mod 11
  =(-2)^87 mod 11
  =(-2)*(-2)*(-2)^85 mod 11
  =4*(-2)^5^17 mod 11
  =4*(32)^17 mod 11
  =4*(-1)^17 mod 11
  =4*(-1)mod 11
  =(-4)mod 11
  =7
• 9 years agoHelpfull: Yes(9) No(5)
• 30^72^87=30^72*30^72*30^72*30^72*30^72*.........87 times.
  first of all we have to find out the reminder for 30^72/11
  euler number of 11 is 10.
  rem(30^70/11)=1
  remaining 30*30/11=8*8/11=9
  
  so....
  9*9*9*9*9*.....87times.
  ==>we have to finout the reminder of 9^87/11
  rem(9^80/11)=1
  remaining 9^7/11=81*81*81*9/11=4*4*4*9/11=81/11=4
• 9 years agoHelpfull: Yes(8) No(2)
• rem = 5
  72^87 unit digit is 8
  
  so
  30^8/11
  (-3)^8/11
  81*81/11
  4*4/11
  16/11
  =5
  
• 9 years agoHelpfull: Yes(5) No(5)
• answer will be 7
  30^72^87/11
  
  8^6^10/11
  
  4^10/11
• 9 years agoHelpfull: Yes(2) No(2)
• answer is 4
• 9 years agoHelpfull: Yes(2) No(0)
• As φ(11) = 10
  => 3010k = 1mod11
  So 7287 is to be written in the form of 10k + a.
  Now unit digit of 7287 = 10k + 8
  => 3072^87 = 30(10k + 8) = 3010k *308=308mod11 = (-3)8mod11 [33-3=30. So (-3)8 ]
  = 33mod11 [ 35 = 1mod11]
  = 27mod11 = 5mod11.
  ans is 5
• 9 years agoHelpfull: Yes(1) No(0)
• Fermat little theorem says, (a^(p−1))/p remainder is 1.
  ie., 30^10 or 8^10when divided by 11 remainder is 1.
  The unit digit of 72^87 is 8 (using cyclicity of unit digits)
  So 72^87 = 10K + 8
  (30^(10K+8))/11=(30^10)^K.30^8/11=1k*30^8/11
  8^8/11=2^24/11=(2^5)^4*2^4/11=16/11=5
• 8 years agoHelpfull: Yes(1) No(0)
• Answer is 1
• 9 years agoHelpfull: Yes(0) No(3)
• remainder is 1
• 9 years agoHelpfull: Yes(0) No(0)
• (30^72)^87%11=30^6264%11;
  =(3*11-3)^6264%11
  =3^6264%11; (since power is even so by formula (a*x-1)^n%a=1 if n is even)
  =3^4 * 3^6360%11
  =(11*7+4)%11 * (3^5)^1272%11
  =4%11 * 243^1272%11
  =4%11 * (11*22+1)1272%11
  =4
• 9 years agoHelpfull: Yes(0) No(0)
• https://www.wolframalpha.com/input/?i=%2850%5E56%5E52+%29mod+11
  
  check here...
• 8 years agoHelpfull: Yes(0) No(0)
• Ans : 5
  remainder should be 5
• 8 years agoHelpfull: Yes(0) No(0)
• we can solve this easily using Euler's Theorem
  
  x^e = x^(emod phi(n) )(mod n)
  
  where ,
  phi(n) = Euler Totient Function
  
  if n can be prime factorised as follows:
  n = p1^k1 * p2^k2 * ........ * pn^kn, where p1,p2,,,,pn are prime factors of n
  then
  phi(n) = n *(1-1/p1)*(1-1/p2)*.......*(1-1/pn)
  
  if n is already a prime number then
  phi(n) = n-1
  
  we will use the above knowledge to solve 30^72^87 mod(11)
  
  now we will go step wise
  
  = 30^72^87 mod(11)
  so we get 30^k (mod 11) where k =72^87 mod(phi(11))
  now k = 72^87 mod(10) as 11 is prme so phi(11) =10
  k =2^87 mod(10)
  which can be again broken down to k = 2^k1 , where k1 = 87 mod(phi(10))
  so prime factors of 10 = 2 ,5
  phi(10)=10 *(1-1/2)*(1-1/5)
  = 4
  therfore k1 =87mod4
  k1 =3
  k =2^k1mod(10)
  = 2^3mod(10)
  k=8
  Now 30^kmod(11)
  =(30mod(11))^8 mod(11)
  =8^8mod(11)
  = 64mod(11) * 64mod(11)*64mod(11) mod(11)
  
  =9 * 9*9*9 mod(11)
  =6561mod(11)
  =5
  
  so remainder = 5
  
  
• 7 years agoHelpfull: Yes(0) No(0)