Find Remainder when 32^32^32 is divided by 9.

• To calculate remainder for a base having 2 exponents, You can follow the below shortcut method...
  Remember 32^32^32 is not equal to 32^(32*32) ...
  
  Here 32^32^32 % 9
  (9*4-4^32^32)%9
  (-4)^32^32
  
  Base 4 nd divisor 9 are co-primes... So we can use Euler's Theorem...
  
  Euler's Theorem :
  *****************
  R[M^N % P] ... If M nd N are co- primes , them We can use this theorem...
  We have to calculate E(N) = N*(1-1/a)*(1-1/b)...
  For example if i want to calculate E(15),
  15 = 3^1*5^1
  Then E(15) = 15*(1-1/3)*(1-1/5) => E(15) = 8
  Similarly 9=> 3^2
  E(9) = 9*(1-1/3) = 6
  Then R[M^E(N) % P] = 1 [This is the Euler's Form]
  
  So now we have to calculate (-4)^32^32 % 9
  If we have 2 exponents (R[M^N^O] % P], this is the procedure
  Calculate N^O % E(N) = x
  Then Calculate M^x % P
  
  So 32^32 % 10 => [10*3 + 2]^ 32 % 6
  2^32 % 6
  We can use cyclic rem. theorem here...
  2^1 % 6 = 2
  2^2 % 6 = 4
  2^3 % 6 = 2 [Again]
  2^4 % 6 = 4
  So cyclicity = 2 [upto 2nd power]So rem ll be 2k+n
  So 32 => 2*15+2 [k=15,n=2]
  If n=1, rem=2
  If n=2, rem=4
  So rem[2^32 % 6] = 4
  
  Now Calculate (-4)^4 % 9
  => 256 % 9
  = 4%9
  Hence rem is 4
• 10 years agoHelpfull: Yes(26) No(2)
• Am sorry...
  32^32%10=(10*3+2)^32%6 .... Its a mistake...
  
  If we have 2 exponents (R[M^N^O] % P], this is the procedure
  Calculate N^O % E(N) = x
  Then Calculate M^x % P
  
  So 32^32 % 6 => [6*5 + 2]^ 32 % 6 [Here E(N) = 6... N^O = 32^32]
  Rem[32^32 % 6] = 4
  
  And Yes 32=> 2^5
  So E(32) = 32*(1-1/2) = 16
• 10 years agoHelpfull: Yes(5) No(1)
• 1 will be the remainder
  
  First of all we have to find out the unit digit of 32^32 , which is 6
  It means (32^6)/9 will give us the remainder.
  
  Now, (32^6)/9 = (9*3+5)^6 /9
  = (5)^6 / 9
  = {(5)^3}^2 / 9
  = (125)^2 / 9
  = (9*14-1)^2 / 9
  = (-1)^2 / 9
  = 1 / 9
  = 1
• 10 years agoHelpfull: Yes(2) No(3)
• 32^1 % 9 = 5
  So (5^32^32) % 9
  Euler value of(9) = 6
  ((5^6)^5 * 5^2)^32
  5^64 % 9
  E(9) = 6
  (5^6)^10 * 5^4 % 9
  625 % 9 = 4
  
  Ans : 4
• 10 years agoHelpfull: Yes(2) No(1)
• Saraswathy answer is 100 percent correct.
  Can you please explain the method to solve....?
  How Euler Value of (9) is 6 and how u put the values in form:
  ((5^6)^5 * 5^2)^32
  
  Plzzz....and thx for ur reply !!
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• Euler's Theorem :
  *****************
  R[M^N % P] ... If M nd P are co- primes , them We can use this theorem...
• 10 years agoHelpfull: Yes(0) No(1)
• @Saraswathy
  Pls explain the statementa given below
  32^32%10=(10*3+2)^32%6
  
  I mean E(32)=32*(1-1/2)=16
  Right?
  I am confused.
• 10 years agoHelpfull: Yes(0) No(0)
• Thanks Saraswathy for ur detailed explanation....thank u very much !! :)
• 10 years agoHelpfull: Yes(0) No(0)
• 32^32^32/9
  32/9=REM 5
  5^32^32/9
  5/9=5;25/9=7;125/9=8;5^4/9=4;5^5/9=2;5^6/9=1;5^7/9=5 REPEATING
  32^32/6->32/6=2(REM)
  2^32/6->2/6=2;4/6=4;8/6=2 REPEATING
  32/2=0(REM);
  2^32/6=1;
  5^1/9=5
  ANSWER 5
• 9 years agoHelpfull: Yes(0) No(0)