find remainder when 33^34^35 divided by 7 ?

• (33)^(34)^(35)/7
  (28+5)^34^35/7 according to remainder theorm remainder will be 5^34^35/7
  34^35 will be a even number
  let write this in the form of 3
  (34^35)/3=(33+1)^35/3 remainder will be 1.
  so 34^35 can be written as 3k+1 (k can be anything doesnt matter)
  as 34^35 is even so 3k+1=even=>k is odd
  5^(3k+1)/7=(125^k)*5/7=(126-1)^k(0dd)*5/7
  so remainder -1*5=-5
  so make the remainder +ve 7-5=2answer
• 9 years agoHelpfull: Yes(37) No(9)
• this method is valid only for consecutive numbers..
  
  rem((33^34^35)/7) = rem [{rem(33/7)+rem(34/7)+rem(35/7)}/7]
  = rem [ (5+6+0)/7]
  =rem [11/7]
  =4
  here rem== remainder
  
• 9 years agoHelpfull: Yes(18) No(16)
• Remainder is 2
  
• 9 years agoHelpfull: Yes(12) No(6)
• 33^34^35 is 33^1190. Now rem( 33^1190)/7= rem( 5^1190)/7=rem (-2^1190)/7
  Now, -2^6 when divided by 7 leaves rem 1.
  Therefore the said prob can be written as rem(-2^1188)/7*rem(-2^2)/7= 1*4=4
  Rem= 4
• 9 years agoHelpfull: Yes(11) No(13)
• correct solution is 2
  33^34^35/7=(-2)^4^5/7=(-2)^4/7=2
• 9 years agoHelpfull: Yes(6) No(4)
• ans is 2 if any doubt then check on == https://www.wolframalpha.com/input/?i=%2833%5E34%5E35%29mod7
  
• 9 years agoHelpfull: Yes(4) No(2)
• 33^34^35 = 33^1190 = (28+5)^1190 = rem of (5)^1190 = rem of (25)^595 = rem of (21+4)^595 = rem of (4)^595= rem of (4^594) * 4 = rem of (64^198) * 4= rem of (63+1)^198 * 4 = rem of 1^198 * 4 = rem of 1*4 = 4
• 9 years agoHelpfull: Yes(3) No(6)
• (33^34^35)/7 can be written as (-2)^34^35/7 => 2^34^35/7
  since 2^6 gives remainder of 1 when divided by 7.
  try to write 34^35 in (6*k+r) form,
  34^35, when divided by 6, gives remainder of 6.
  so 34^35 can be written as (6*k+4),
  thus, 2^34^35 can be written as 2^(6*k+4).
  thus the remainder will be (2^4)%7=2.
  
  hope this helps.
  
• 9 years agoHelpfull: Yes(3) No(2)
• watch this remainder theorem concept in youtube for better understanding ...
  http://www.youtube.com/watch?v=XEVuGq89tE0
• 9 years agoHelpfull: Yes(3) No(1)
• (33)^(34)^(35)/7
  
  that means we can write (34)^(35)=N
  so (35)^N/7=(5)^N/7
  case 1:
  5^1/7=5
  5^2/7=4
  5^3/7=-1
  so we can get 1 at 5^6/7=1(remainder)
  so cycle of 6 so N/6=34^35/6=(4)^35/6
  we can also write as 4*4^34/6=2*(4)^34/3=2*(1)^34/3=2 or we have cancel 2 before so multiply
  by 2=2*2=4
  so the value of N is 4
  and now (5)^4/7=-5 or 2 answer
  
  
• 9 years agoHelpfull: Yes(3) No(1)
• remainder is 2
• 9 years agoHelpfull: Yes(2) No(1)
• (33^34^35)/7= (33^34^0)/7.... since 35/7=0, now 34^0=1
  now, (33^1)/7= 33/7 so remainder will be 5
  
• 9 years agoHelpfull: Yes(2) No(2)
• Ans is 0.
  
  [(-2)*(-1)*0]/7
  => 0/7
  => 0.
• 9 years agoHelpfull: Yes(2) No(0)
• sorry, in the above solution, 34^35 when divided by 6, gives remainder 4.
• 9 years agoHelpfull: Yes(1) No(1)
• remainder is 2
  solve with eulers theorem
  
• 9 years agoHelpfull: Yes(1) No(1)
• rem of (33^34^35)/7:
  
  34 * 35=1190;
  
  
  (33^1190)%7= (35-2)^1190%7
  = (2^1190)%7
  = (2^2*2^1188)%7
  = (4*(2^3)^396)%7
  = (4*(7+1)^296)%7
  =4%7
  =4
• 9 years agoHelpfull: Yes(1) No(1)
• 4
  33^34^35=4
  
• 8 years agoHelpfull: Yes(1) No(0)
• 33^34^35/7 =-2^-1^0(using negative remainder theorem)=2
• 9 years agoHelpfull: Yes(0) No(2)
• remainder is 2.
  
• 9 years agoHelpfull: Yes(0) No(0)
• 2 is d ans
• 9 years agoHelpfull: Yes(0) No(0)
• (33)^(34)^(35)/7
  now 34^35 divided by 3 remainder is 1
  (98+1)^(34^35)/3
  1^(34^35)/3
  1^(3k+1)
  1^3k+1
  1+1=2 ans
• 9 years agoHelpfull: Yes(0) No(0)
• Remainder of (33,34,35) when divided by 7 are (-2,-1,0) .now multiply the remainders except 0 i.e. 2. So, 2 is the remaind.
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• The remainder is 2
  35% 7 = 0
  34% 7 = 6
  33 % 6 = 5
  5^6
  30%7
  2
  
• 9 years agoHelpfull: Yes(0) No(1)
• unit digit of 33^34= 9
  unit digit of 9^35=9
  so,
  9/7 gives the remainder as 2
  so ans is 2
• 9 years agoHelpfull: Yes(0) No(0)
• we can solve this easily using Euler's Theorem
  
  x^e = x^(emod phi(n) )(mod n)
  
  where ,
  phi(n) = Euler Totient Function
  
  if n can be prime factorised as follows:
  n = p1^k1 * p2^k2 * ........ * pn^kn, where p1,p2,,,,pn are prime factors of n
  then
  phi(n) = n *(1-1/p1)*(1-1/p2)*.......*(1-1/pn)
  
  if n is already a prime number then
  phi(n) = n-1
  
  we will use the above knowledge to solve 33^34^35 mod(7)
  
  now we will go step wise
  
  =33^34^35 mod(7)
  so we get 33^k (mod 7) where k =34^35 mod(phi(7))
  now k = 34^35 mod(6) as 7 is prme so phi(7) =6
  k =4^35 mod(6)
  which can be again broken down to k = 4^k1 , where k1 = 35 mod(phi(6))
  so prime factors of 6 = 2 ,3
  phi(6)=6 *(1-1/2)*(1-1/3)
  = 2
  therfore k1 =35mod2
  k1 =1
  k =4^k1mod(6)
  = 4^1mod(6)
  k=4
  Now 33^kmod(7)
  =(33mod(7))^4 mod(7)
  =5^4 mod(7)
  = 25mod(7) * 25mod(7) mod(7)
  =4 * 4 mod(7)
  =16mod(7)
  =2
  
  so remainder = 2
  
  
  
  
  
  
• 8 years agoHelpfull: Yes(0) No(0)
• Ans : 2
  sol : 33^34^35/7;
  = (35-2)^34^35/7
  =(-2)^34^35/7=(2)^34^35
  =((2^3)^11^35*2^1^35)/7
  = ((7+1)^11^35*2)/7 formula (a+1)^n=1
  =2
• 8 years agoHelpfull: Yes(0) No(0)
• ans is 243
• 8 years agoHelpfull: Yes(0) No(0)
• 33/7=5
  
  5^34^35/7
  
  cycle
  5^1/7=5
• 8 years agoHelpfull: Yes(0) No(0)