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

• we can use remainder theorem when consecutive no. are there...so here we cosider it as 32*33*34 but as 33 divisible its not required so 32*34 so taking -ve and +ve remainder i.e. (-1)*(1) = -1 now the remainder is
  11-1 = 10.(bcoz when the remainder is -ve we subtract it from no. dividing) so ans is 10.
• 10 years agoHelpfull: Yes(16) No(4)
• 32^33^34 = 32^1122
  32^1122/11 by using remainder theorem ans will be 1
• 10 years agoHelpfull: Yes(11) No(5)
• @ RUDRA
  ans is 10
  use remainder theorem man
  (32^33^34)11(when consecutive such as 51*52*53 or 22*23*24 num are qiven)
  take them as product (32*33*34)/11
  as 33 is divisible by 11 leave it...
  32 when divided by 11 leaves remainder -1(negative remainder)
  similarly 34 when divided by 11 leaves remainder 1(+ve remainder)
  (-1*1)/11 = -1/11
  as we got negative term in numerator subtract it from denominator
  so remainder = 11-1=10
  (if you got positive term in numerator then leave it....it will be the remainder)
  
• 10 years agoHelpfull: Yes(8) No(0)
• (32^33^34)/11=(-1^33^34)...33^34 is always odd so -1...ie -1+11=10...so remainder is 10
• 10 years agoHelpfull: Yes(6) No(6)
• use remainder therorm
  since 33 is divisible by 11 remainder is 0
  32 is divided by 11 =-1
  33 divided by 11= 1
  1*(-1)=-1
  11-1=10(since the remainder is -ve)
  so ans-10
• 10 years agoHelpfull: Yes(6) No(0)
• remainder is 1.
  
• 10 years agoHelpfull: Yes(3) No(6)
• (33-1)^33^34/11
  from reminder theorem
  1^33^34/11=1
• 10 years agoHelpfull: Yes(1) No(1)
• 32^33^34/11=10^33^34/11=(-1)^33^34/11=1/11
  thus remainder is 1
• 10 years agoHelpfull: Yes(1) No(1)
• @SARIKA plzzzz explain in detail........
• 10 years agoHelpfull: Yes(0) No(0)