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

• 30^72^87 mod 11
  30^72^87 mod 11
  
  Euler value of 11 = 10
  72^87 mod 10
  (7*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 [Cyclicity is 4... Rem = (4k+3)rd value]
  2^87 mod 10 = 8
  
  Now 30^8 mod 11
  (11*2+8)^8 mod 11
  8^8 mod 11
  2^24 mod 11
  Euler value of 11 = 10
  (2^10)^2 * 2^4 mod 11
  16 mod 11 = 5
  
  Ans : 5
  
• 9 years agoHelpfull: Yes(19) 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=(1^k).(30^8)/11
  (8^8)/11=(2^24)/11=16/11=5(remainder)
• 9 years agoHelpfull: Yes(2) No(0)
• N = a^x*b^y*c^z* ... etc
  Euler value of N is => N*(1-(1/a)) * (1-1/b) * (1-1/c)..etc
  Lets take a number 10
  10 => 2*5
  E(10) = 10*(1-1/2)*(1-1/5) => 4
  Lets take 21
  21 => 3*7
  E(21) = 21*(1-1/3)*(1-1/7) => 12
  
  Hope you understand this @sudhir sharma
  
• 9 years agoHelpfull: Yes(2) No(0)
• Lets take an example...
  If we want to calculate remainder of 2^17 divided by 10
  Can u say the value of 2^16 without calculator? ... Its impossible na...
  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
  2^6 mod 10 = 4
  2^7 mod 10 = 8
  2^8 mod 10 = 6
  [Note here u dont need to calculate 2^8... 2^8 is 2^7*2^1...
  So remainder of 2^7 * 2^1=8*2 = 16 mod 11 = 6]
  
  If u observer carefully, after 4th power, the remainder is again 2,4,8,6... Again it goes on like 2,4,8,6...[This is cyclicity]
  So remainder of 2^16 ll be 4k+n since 1 cycle get over upto 4th power...
  17 => 4k+n => 4*4 + 1
  So remainder of 2^17 mod 10 ll be 2^n = 2^1 = 2...
  Hope u understand this... @Suraj
  
• 9 years agoHelpfull: Yes(2) No(0)
• saraswathy, how u calculate euler's value?
• 9 years agoHelpfull: Yes(0) No(0)
• Thanks @ Saraswathy
• 9 years agoHelpfull: Yes(0) No(0)
• how saraswaty???i could not understand...
• 9 years agoHelpfull: Yes(0) No(0)
• What u cant understand @Suraj jha...
• 9 years agoHelpfull: Yes(0) No(0)
• what about cyclicity???i could not understand...
• 9 years agoHelpfull: Yes(0) No(0)
• if you dont have any problem....can u please tell the answer on the email id i just sent to you...
• 9 years agoHelpfull: Yes(0) No(0)
• a bit...problem remains
• 9 years agoHelpfull: Yes(0) No(0)