Remainder when 30^72^87 divided by 11

• if its 30^(72^87)/11
  
  72^87 gives unit digit 8 so it can be written as (10k + 8)
  
  30^(10k+8)/11
  => 30^10k * 30^8 / 11
  => (-3)^10k * (-3)^8 / 11
  => (3^5)^2k * (3)^5* 3^3 / 11
  => (11*22+1)^2k * (11*22+1)^5 * 27 / 11
  => 1^2k * 1^5 *27 /11
  => 27/11
  => rem = 5
• 10 years agoHelpfull: Yes(33) No(8)
• if its (30^72)^87/11
  
  (30^72)^87/11
  => 30^6264/11
  => (11*3-3)^6264/11
  => (-3)^6264/11
  => ((-3)^2)^3132/11
  => 9^3132/11
  => (11-2)^3132/11
  => 2^3132/11
  => (2^5)^626* 2^2 /11
  => (33-1)^626 * 4 /11
  => 4/11
  => rem = 4
• 10 years agoHelpfull: Yes(25) No(11)
• dear rakeshbhaya,the answer given in arun sharma for this problem is 5
• 10 years agoHelpfull: Yes(10) No(1)
• please notify me where i am wrong according to me ans is 7
  here is my calculations
  according to remainder theorem
  30*72*87/11 --R->(-3)(6)(-1)/11 =(18)/11-R-->7ANS
• 10 years agoHelpfull: Yes(5) No(6)
• i thimk it will be 7
  
• 10 years agoHelpfull: Yes(3) No(3)
• : Fermat little theorem says, ap−1p 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) Click here
  So 72^87 = 10K + 8
  30^(10K+8) / 11=(30^10)K.30^8/11=1k.30811
  8^8/11=2^24/11=(2^5)4.2^4/11=16^11=5
  
• 10 years agoHelpfull: Yes(2) No(2)
• 11)30(2
  22
  ---------
  08............REMAINDER
  WHAT EVER THE VALUE OF 30^X TAKE ONE 30 FROM 30^X FOR DIVDING BY 11
  
• 10 years agoHelpfull: Yes(1) No(0)
• HCF ( 30 , 11 ) is 3.
  where phi(11)=(1-1/11)*11=10
  
  so, 7287 = 10k + 8
  => 3072^87 = 30(10k + 8) = 308mod11 = (-3)8mod11 = 33mod11 [ 35 = 1mod11]
  = 27mod11 = 5
  
• 10 years agoHelpfull: Yes(1) No(1)
• 5 will be the correct Answer.
  Because when u calculate unit digit of 72^87 it will give 8 as unit digit.
  Therefore when 30^8 is divided by 11 , using Remainder Theorem we will get
  5 as remainder.
• 10 years agoHelpfull: Yes(0) No(4)
• @rakesh why answer is not equal to 4 ? could you please explain ?
• 9 years agoHelpfull: Yes(0) No(0)
• (30^72)^87 this can be written as : 30^6264
  
  and 30^6264 can be written as : (30^2)^3132=900^3132
  
  900^3132/11= (11* 80+ 10)^3132/11
  =10^3132/11
  =(11-1)^3132/11
  =1 (since ,power is positive )
• 9 years agoHelpfull: Yes(0) No(1)