Q. What will be the remainder when (32^32^32) is divided by 7.

• a^b^c is a^(b*c)
  hence 32^32^32 is 32^(1024)
  32^1024 mod 7 is= (32^2) ........512 times mod 7
  by the modulus formula ab mod x can be rewritten as a mod x * b mod x
  hence 32^2 mod 7 * 32^2 mod 7 *....................(512 times)
  2*2*.....(512 times) mod 7
  2^512 can be rewritten as (2^3)^170 * 4 mod 7
  2^3 mod 7 is 1 and 1^170 is 1
  1*4 is 4
  hence answer is 4.
  
  
• 10 years agoHelpfull: Yes(2) No(1)
• rem(32/7)=4
  rem(4^2*(4^3)^10^32)/7= rem[(16/7)*rem((64/7)^10^32)= 2*1= 2
  
  the remainder will be 2
• 10 years agoHelpfull: Yes(1) No(0)
• ans is 1.
  first, we can reduce question to 32^64.
  now 32*32*32........ (64 times)/7
  dividing single 32 by 7 will give remainder 4.
  now we get, 4*4*4.........(64 times)/7
  or 16*16*16.............(32 times)/7
  after dividing 16 by 7 we get 2 rmainder thus
  it becomes, 2*2*2......(32 times)/7
  or 8*8*8.........(8 times)/7
  after dividing 8 by 7 we get remindr 1.
  thus, 1*1*1......(8 times)/7
  finally we get, 1/7=1.
  remainder is 1.
• 10 years agoHelpfull: Yes(0) No(2)
• ans is 4.
  32^1024
• 10 years agoHelpfull: Yes(0) No(1)
• 32^32^32/7==4^32^32/7 now cyclicity of 7 ==3(4,1,2,again 4) now 32^32 /3==-1^32=1.so 4^1=4 ans
  
• 10 years agoHelpfull: Yes(0) No(2)
• ans is 4 becauslate after calcu
• 10 years agoHelpfull: Yes(0) No(0)