Find the remainder when 32^33^34 is divided by 11.

• answer=1
  32^33^34= (33-1)^33^34/11= (-1)^33^34/11= (-1)^even number(33*34) /11
  = 1/11= 1(remainder)
• 11 years agoHelpfull: Yes(30) No(18)
• answer=10
  32^33^34= (33-1)^33^34/11= (-1)^33^34/11= (-1)^odd no.(as last digit of 33*34 is odd)
  = -1/11= -1 or 11-1=10(remainder)
• 11 years agoHelpfull: Yes(30) No(20)
• 32^33^34/11
  can be written as
  32^N/11 where N= 33^34
  Now, term reduces to 10^N/11
  Now find the cyclicity:-
  10^1/11 = 10 ( remainder)
  10^2/11= 1 remainder)
  10^3/11 = 10 ( remainder)
  10^4/11= 1 (Remainder)
  Hence it has cyclicity of 2
  now question reduces to
  N/2
  Now find the remainder
  of 33^34/2
  Remainder= 1
  First term of the cycle will be the answer
  Hence Final answer is 10
  
• 11 years agoHelpfull: Yes(10) No(4)
• 10
  (33-1)^odd number(as odd^even is always equal to odd)/11
  ==-1/11
  = 10
• 11 years agoHelpfull: Yes(7) No(7)
• answer will be 10 .
• 11 years agoHelpfull: Yes(3) No(2)
• ans 0.
  32^33^34/11 write it as 32^n/11 remainder in this case is calculated first.it can be written as 10^n/11 or (-1)^n/11 so remainder will be -1 or 10.its cycle will be 2.now 33^34/2 gives 1 remainder so the whole remainder will be 0.
• 11 years agoHelpfull: Yes(1) No(18)
• 10 is ans.
• 11 years agoHelpfull: Yes(1) No(4)