30^72^87 divided by 11 gives reminder ?

• Fermat little theorem says, [a^(p−1)]/p remainder is 1.
  30^10 ,, when divided by 11 remainder is 1.
  The unit digit of
  72^87 --> 2^87 --> (2^4)^21 * 2^3
  now unit place of (2^4)^21 * 2^3 --> 6 * 8 = 48 ---> unit place -> 8
  So, 72^87 = 10K + 8
  => 30^72^87 = 30^(10K+8)
  30^(10K+8)/11=[(30^10K) * (30^8)]/11= [(30^10K)/11] * (30^8)/11
  =1k * (30^8)/11
  now, (30^8)/11= [(2 * 11 +8)^8]/11
  (8^8)/11=(2^24)/11=[(2^10)^2 * (2^4)]/11=16/11=5 remainder
• 9 years agoHelpfull: Yes(7) No(5)
• http://www.quora.com/What-is-the-remainder-of-30-72-87-when-divided-by-11
• 9 years agoHelpfull: Yes(4) No(2)
• 30^72^87=30^6264/11
  =(33-3)^6264/11
  using binomial theorem,
  the last term in the expansion is (-3)^6264/11
  =((-3)^2)^3132/11
  =9^3132/11
  =(11-2)^3132/11
  the last term in the expansion is
  =(2^(5*626)+2)/11=(32^626)*4/11
  ((33-1)^626)*4/11=4/11 remainder is 4
  
• 9 years agoHelpfull: Yes(2) No(8)
• 30^72^87=( 10*3)6264
  =10^6264/11 * 3^6264/11
  = 1*(3^10)^626*3^4/11
  =3^4/11=rem 4
• 8 years agoHelpfull: Yes(1) No(1)
• 5 is the correct answer
  
• 9 years agoHelpfull: Yes(0) No(1)
• 30/11=8=2^3
  (2^3)^72^87/11 =((2)72*3)^87/11=(2^216)^87/11=((2^5)43.2)^87)/11=((32)^(43*87).2^87)/11
  =((-1)^odd.(2^5)^17.2^2)/11=((-1)(-1).4)/11=4/11
  from remainder theorem
  hence remainder=4..
• 8 years agoHelpfull: Yes(0) No(0)
• 
  ans. is 5
  
  
  
• 8 years agoHelpfull: Yes(0) No(0)
• Ans 5
  72^87=2^87=8 (87/4 gives remainder 3 and 3rd cycle of 2 is 8)
  30^8=(11*2+8)^8
  =(11*2)^8+8^8
  =(11-3)^8
  =3^8
  =6561/11=5 (no need to take any variable)
• 8 years agoHelpfull: Yes(0) No(0)