30^72^87 divided by 11 gives remainder?

• 72^87 gives last digit as 8 and it can be written in the form 10*k+8 form
  so qstn reduces to 30^(10*k+8)
  now using fermatt rule 30^10k =1(mod11)and 30^8/11 gives 5 using remainder theorem so remainder is 5
• 11 years agoHelpfull: Yes(19) No(17)
• 1 may be the answer
• 11 years agoHelpfull: Yes(3) No(6)
• ans will be 3.
• 11 years agoHelpfull: Yes(3) No(3)
• remainder is coming as 4
• 11 years agoHelpfull: Yes(3) No(4)
• 3
  [30/11]=rem(8)
  8/11=rem(8)
  8^2[8*8/11]=rem(9)
  8^3[9*8/11]=rem(6)
  8^4[6*8/11]=rem(4)
  8^5[4*8/11]=rem(10)
  8^6[10*8]=rem(3)
  ........
  .......
  .......
  8^10[56%11]=rem(1)
  8^3[8*9/11]=rem(6)
  8^4[6*8/11]=rem(4)
  ........continue u will get cyclcity as 10
  supos(72^87)=N
  N/10=72^87/10
  remainder coming 6 ,now 6 means 6th part of the cycle which is 3
  
• 11 years agoHelpfull: Yes(1) No(1)
• remainder is 4
  
• 11 years agoHelpfull: Yes(0) No(1)