What's the remainder if 32^32^32 is devided by 9 ?
A)4
B)7
C)1
D)2

• 32^32^32=2^5120
  use pattern method
  1) 2^1/9=2
  2) 2^2/9=4
  3) 2^3/9=8
  4) 2^4/9=7
  5) 2^5/9=5
  6) 2^6/9=1
  after 6 steps pattern is repeating
  7) 2^7/9=2
  
  divide 5120 and see whats the remainder rem=8 after 6 steps 2 steps gives 4
  so ans is a)4
  
• 11 years agoHelpfull: Yes(39) No(4)
• can anyone explain how 32^32^32= 2^5120... ??
  
• 11 years agoHelpfull: Yes(12) No(2)
• 32^32^32=2^5120
  >> 2^2 * [2]^5118
  >> 4* [2^3]^1706
  >> 4* [8]^1706
  >> 4* [9-1]^1706 using ((a-1)^n)/a= 1 if n is even, else a-1.
  >> 4* [-1]^1706, because power is even, evaluation of brackets will give 1.
  thus 4*1=4 Ans
• 11 years agoHelpfull: Yes(6) No(2)
• since unit place 2^2 = 4 so 2^32 = 4
  ||ey 2^4 = 16/9 = 7 remainder
• 11 years agoHelpfull: Yes(5) No(3)
• ans is D)2
  
• 11 years agoHelpfull: Yes(2) No(3)
• u r right nishant the ans is 4 my mistake
  
• 11 years agoHelpfull: Yes(2) No(0)
• see the unit place..
  2^2^2= 16
  16/9 = 7
  Ans=>7
• 11 years agoHelpfull: Yes(1) No(4)
• what is the ans??pls give someone the right xplanation...my ans is 7
• 11 years agoHelpfull: Yes(1) No(4)
• 32^1=32
  32^2=1024
  32^3=----68
  32^4=-------76
  32^5=------32
  so after that it goes on repeating like this.....
  thus 32/5 gives two extra.....
  hence last two digits of 32^32=68......
  now 32^32^32=32^68 considering only last two digits....
  now 68/5 gives 3 extra...
  thus it will give 76 at last two places..
  now 76/9 gives remainder..4
  
  ans:- 4
• 11 years agoHelpfull: Yes(1) No(0)
• question can be written as
  32^(32*32*.....32times)/9= 2^(5*32*32....32times)/9 =
  [2^(32*32....32times)]*[2^(32*32......32times)]*....5times/9 (because 5 is there i wrote it 5 times seperately)
  but 2^32*32*.....32times = 2^4n=2;
  therefore
  above big expression = 2*2*2*2*2/9 = 2^5/9 = 2*2^4/9 = 2*2/9 = 4/9 =4
  (4 is the answer).
• 11 years agoHelpfull: Yes(1) No(0)
• right ans will be 4
• 11 years agoHelpfull: Yes(0) No(1)
• sorry,
  2*2^4/9 = 2^5/9 = 2/9.
  so, the answer is 2.
  sorry i have given it as in the above solution of myne as 4.please correct it.
• 11 years agoHelpfull: Yes(0) No(0)
• 32^32^32=
  ((2^5)^32)^32=
  (2^160)^32=
  2^160*32=
  2^5120
  sarab this is the step
• 11 years agoHelpfull: Yes(0) No(0)
• 32^32^32/9
  =32^64/9
  =2^5^64/9
  =2^69/9
  now 2^9=256
  256/9 then remainder=4
  answer=4
• 7 years agoHelpfull: Yes(0) No(0)
• HCF(9,32)=1 so,both are co-prime
  rem[32/9]=5
  so rem[5^f(9)/9]=1
  so f(9)=9*(1-1/3)=>6
  rem[5^6/9]=1
  now , rem[32^32/6]=4
  so , 32^32=6n+4
  so,rem[ 5^(6n+4)/9]=>rem[5^6n/9] *rem[5^4/9]
  =>1 *rem[4/9]=4(ans)
• 7 years agoHelpfull: Yes(0) No(0)