What is the remainder when 32^33^34 is divided by 11?

• 32^33^34/11=32^(33*33*.....34 times)/11
  now remainder of 32/11=10
  therefore 10^(33*33*....34 times)/11 => 10^(n)/(10+1) => 10 [where n=odd bcz any power of 33 is odd]
• 10 years agoHelpfull: Yes(22) No(3)
• Ans is 10
  32=-1 mod 11
  now,
  -1^odd=(-1)^even*(-1) ----(*)
  Here 33^34 is odd (Already said)
  So,
  (-1)^even=1
  so (*) equation is reduced to
  -1 mod 11
  or 10
• 10 years agoHelpfull: Yes(10) No(3)
• 32%11=-1(rem)
  -1^odd = 1 so rem= 1
• 10 years agoHelpfull: Yes(9) No(13)
• plzz explain in detail
• 10 years agoHelpfull: Yes(3) No(2)
• 32^33^34/11=32^(33*33*.....34 times)/11
  now remainder of 32/11=10
  therefore 10^(33*33*....34 times)/11 => 10^(n)/(10+1) => 10 [where n=odd bcz any power of 33 is odd]
• 10 years agoHelpfull: Yes(2) No(1)
• remainder is 1.because 33*34 is even.(-1)to the power even is 1.
• 10 years agoHelpfull: Yes(2) No(2)
• answer:7 by power cycle
• 10 years agoHelpfull: Yes(1) No(3)
• we can multiply the powers since they r small so 33*34=1122
  now if we divide 32 by 11 we get a negative remainder i.e -1 so our equation will be reduced in (-1)^1122/11 now when -1 will have even power it will come to be +1 so we get 1 as our answer .
• 10 years agoHelpfull: Yes(1) No(0)
• 100
  plz
  now remainder of 32/11=10
• 9 years agoHelpfull: Yes(0) No(0)