30^72^87 divided by 11 gives remainder?

• can be solved only knowing remainder theorem
  
  given%11= (30%11)^72^87=8^72^87
  (8^72^87)=(64%11)^36^87=9^36^87
  9^36^87=(81%11)^18^87=4^18^87
  (16%11)^9^87=5^9^87
  when 5 comes it will recur whatever may b the power
• 9 years agoHelpfull: Yes(30) No(2)
• unit digit of 72^87 = 8
  so,72^87 = (10k + 8)
  
  30^(10k+8)/11
  = 3^(10k+8)*10^(10k+8)/11
  = 3^10k * 3^8 * 10^(10k+8)/11
  = (3^5)^2k * 3^5*3^3 * 10^(10k+8)/11
  = (11*22+1)^2k *(11*22+1)*(11*2+5)*(11*1-1)^(10k+8)/ 11
  
  => rem = 1^2k * 1 * 5 *(-1)^(10k+8) = 5
• 9 years agoHelpfull: Yes(28) No(4)
• (22+2^3)^8%11=5
• 9 years agoHelpfull: Yes(4) No(1)
• 30^72^87 /11
  so by the form a^p-1/p the rem is always 1.
  so we can say that 30^10 or 8^10 the rem=1;
  
  the unit digit of 72^87 is 8 by cyclicity of unit digits.
  
  so 72^87 =10L+8
  
  30^(10L+8)/11
  => (30^10)^L *30^8/11
  => 1^L*30^8/11
  => 30^8/11
  => 8^8/11
  => 2^24/11
  => (2^5)4*2^4/11
  => 16/11= 5
  so 5 is rem Ans
• 9 years agoHelpfull: Yes(1) No(4)
• 1st-rem(30/11)=8
  2nd-8^72^87/10
  3rd-euler(11)=10
  4th-rem( (72*87)/10)=4
  5th-8^4/11=2^12/11
  6th-rem(12/10)=2
  7th-rem(4/11)=4
  ans=4
  
  please response if my solution is right or wrong.
  
  
• 9 years agoHelpfull: Yes(1) No(4)
• 72, unit digit is 2..now 2^87=last digit will be 8...30=22+8 ,,22 s divisible by 11...so 8^8=2^24
  now 2^24 mod 11 have to conclude..
  so
  2^2 mod 11=4
  2^4mod 11=(2^2)^2 mod 11=4^2mod 11=5
  2^8 mod 11=(2^4)^2 mod 11=5^2mod11=3
  2^16 mod 11=(2^8)^2 mod 11=3^2 mod 11=9
  2^24 mod 11=(2^8x2^16)mod 11=3x9 mod 11=5 remainder as 27/11=5 remainder...
• 9 years agoHelpfull: Yes(1) No(0)
• can we do like this...
  
  30^6254/11= (33-3)^6254/11 = (-3)^6254/11
  
  (-3*-3)^3127/11 =(9)^3127/11 =((9^3)^1042 *3)/11
  
  (632+1)^1042/11 *3/11= ans 3
• 9 years agoHelpfull: Yes(0) No(2)
• what is remainder therom
• 8 years agoHelpfull: Yes(0) No(0)