Remainder when 33^34^35 divided by 11

• 33^34^35/11
  => (11*3+0)^34^35/11
  => rem(0^34^35) = 0
  remainder = 0
• 10 years agoHelpfull: Yes(60) No(3)
• Correct answer is 2
  Because when when we solve 34^35 it will give 4 as a unit digit.
  Therefore when 33^4 is divided by 7 it will give 2 as a remainder.
• 10 years agoHelpfull: Yes(3) No(7)
• zero , coz 33 itself is a factor of 11
• 10 years agoHelpfull: Yes(3) No(0)
• a/c to remainders theorem 33^34^35=(0)(1)(2)=0
  so ans is zero
  
• 10 years agoHelpfull: Yes(3) No(0)
• i think the answer is zero.33^34^35 will be a multiple of 11
• 10 years agoHelpfull: Yes(1) No(3)
• 0 because 33 divide by 11 gives 0 remainder
• 10 years agoHelpfull: Yes(1) No(2)
• for any prime p,
  (a^p) leaves a remainder of 'a' upon division by 'p'.
  
  here p = 11
  and a = 32
  
  we have 32^(11*3)^34
  and clearly 33^34 leaves a remainder of 0 when divides by 11.
  thus we can write:
  32^(11k) for some integer k.
  
  using fermat's little theorem,
  32 leaves a remainder of 10 upon division by 11
  the answer is 10
• 9 years agoHelpfull: Yes(1) No(0)
• Answer is 0.
  Given that (33^34^35)/11
  =>(33/11)^34^35
  So Remainder is Zero.
• 10 years agoHelpfull: Yes(0) No(1)
• Remainder is '0'
  33^34^35=
  first of all 35/34 remainder is 1 than
  33^1=>33
  al last 33/11=3 and remainder is zero(0).
  
• 10 years agoHelpfull: Yes(0) No(4)
• LET 34^35=X
  33^X/11=(33^(X-1)*33)/11
  11)33^(X-1)*33(3
  33
  -------------------
  00----REMAINDER
  
• 10 years agoHelpfull: Yes(0) No(0)
• HCF ( 33 , 11 ) is 0.
  where phi(11)=(1-1/11)*11=10
  => 50^10k = 1mod11
  So 34^35 is to be written in the form of 10k + a.
  Now unit digit of 34^35 = 4 => 34^35 = 10k + 4.
  => 33^34^35 = 33(10k + 4) mod 11= 33^4 mod11 =0
  
• 10 years agoHelpfull: Yes(0) No(0)