32^32^32/7 what is the remainder?

4^32^32/7 as 32%7=4
now, 4^3/7 gives us remainder 1;
we have to write 32^32 in the form of 4^3n+x where x is the unit digit of 32^32 ie 6
so,
we can re write the problem as
4^3n+6/7
=>4^3n/7 always gives remainder 1
and 4^6/7 gives remainder 1.
so final remainder should be 1...what is wrong in it ? as the ans given is 4.

• 32^32^32 mod 7
  4^32^32 mod 7
  E(7) = 6
  32^32 mod 6
  2^32 mod 6
  2^1 mod 6 = 2... 2^2 mod 6 = 4... 2^3 mod 6 = 2 again...
  Cyclicity is 2... So rem[2^32/6] = 2k+'n'th value...
  32 => 2*16+0..
  2^32 mod 6 = 4
  Now 4^32^32 mod 7 is reduced to 4^4 mod 7
  16*16 mod 7 => 2*2 mod 7
  4 mod 7
  
  Ans : 4
• 10 years agoHelpfull: Yes(13) No(1)
• Sorry @aniruddha... I didnt understand what you have done...
  I think it might be applicable for (32^32)^32...
  Remember (32^32)^32 is not equal to 32^32^32...
  (2^3)^2 = 2^6 = 64
  2^3^2 = 2^9 = 512...
  E(7) is nothing but Euler value of 7...
  E(N) can be calculated by N*(1-1/a)*(1-1/b)*(1-1/c)... where a,b,c are prime factors of N...
  So 7 => 7^1 [a^x]
  So E(N) = 7*(1-1/7) = 6
• 10 years agoHelpfull: Yes(3) No(0)
• @saraswathy what does E(7) stand for in your solution? Can you tell me what is wrong in the above mentioned process that I have done
• 10 years agoHelpfull: Yes(1) No(0)
• @ saraswathy can you give me the link where a^b^c/7 or any no. type problems are explained thoroughly with the theory behind it?? It will be a great help if u can give me the link. I cant understand why we follow these steps.
• 10 years agoHelpfull: Yes(0) No(0)
• 32^32 its unit place is 6
  now (2^5)32^32=2^2(7+1)^last digit will be 6(even number)---equationn no. (1)
  we know that
  (aX+1)^n/X= 1 remainder
  if we devide equation no (1) by 7 then we get
  4(1)/7=4
  
• 10 years agoHelpfull: Yes(0) No(1)
• (32^ 32^32) / 7 = (4^32^32) / 7 .......................(as 32/7 rem is 4)
  now
  search for cyclicity of 4/7... in this way.......
  4/7 = (rem) 4
  4^2 / 7 = (rem) 2
  4^3 / 7 = (rem) 1............hence cycle of 3
  
  now divivide 32^32 / 3
  same way as above..
  32^32 / 3 => 2^32 / 3
  2 / 3 = 2
  2^2 / 3 = 1 cycle of 2
  
  32 / 2 = 0(rem)
  
  0 means last term of cycle..that is 1 in 2^2 /3 and then 1st term of cycle in 4 / 7
  that is 4 as remainder
  
• 9 years agoHelpfull: Yes(0) No(0)