(32^32^32)/9 will leave a rimender?????
1- 4
2- 7
3- 1
4- 2

• 32 = 2⁵
  (2⁵)³² = 2¹⁶⁰
  (2¹⁶⁰)³² = 2⁵¹²⁰
  You are looking for the remainder of 2⁵¹²⁰ ÷ 9.
  
  2¹ mod 9 = 2
  2² mod 9 = 4
  2³ mod 9 = 8
  2⁴ mod 9 = 7
  2⁵ mod 9 = 5
  2⁶ mod 9 = 1
  2⁷ mod 9 = 2
  2⁸ mod 9 = 4
  2⁹ mod 9 = 8
  2¹⁰ mod 9 = 7
  2¹¹ mod 9 = 5
  2¹² mod 9 = 1
  ...
  Note that the pattern repeats, and when the exponent is divisible by 6, the remainder is 1.
  5120 = 6(853) + 2, therefore
  2⁵¹²⁰ mod 9 = 4
• 9 years agoHelpfull: Yes(35) No(0)
• ANS IS : 4
  =(32^32^32)/9
  =(2^5)^32
  =(2^160)^32
  =(2^5120)/9
  2¹ mod 9 = 2
  2² mod 9 = 4
  2³ mod 9 = 8
  2⁴ mod 9 = 7
  2⁵ mod 9 = 5
  2⁶ mod 9 = 1
  2⁷ mod 9 = 2
  2⁸ mod 9 = 4
  2⁹ mod 9 = 8
  2¹⁰ mod 9 = 7
  2¹¹ mod 9 = 5
  2¹² mod 9 = 1
  LOOKING FOR CYCLIC REMAINDER THEOREM
  5210/9=REMAINDER IS 8
  8TH POSITION VALUE IS 4
  
• 9 years agoHelpfull: Yes(12) No(0)
• 32^32^32=(2^5)^32^32=2^5120.
  2^5120/9
  =((2^3)^1706*2^2)/9
  =((9-1)^1706*2^2)/9
  =(-1)^1706*4
  =4
• 9 years agoHelpfull: Yes(9) No(0)
• the last 2 digits of 2^10 will always be 76.
  so the last 2 digits of 2^5120=76.
  76/9 = remainder is 4
• 9 years agoHelpfull: Yes(1) No(0)
• please answer to this question??
• 9 years agoHelpfull: Yes(0) No(1)
• answer will be 4
• 9 years agoHelpfull: Yes(0) No(3)
• ans will be 4
  
• 9 years agoHelpfull: Yes(0) No(2)
• 32/9=5(rem)
  5^32^32/9
  
  5/9=5
  25/9=7
  7*5 / 9=8
  
  5^6 / 9 =1 rem= 1
  
  then 32^32/6= 2^32/6
  2/6=2
  4/6=4
  8/6=2
  16/6=4
  odd even funda
  4th term of cycle that is 5^4/9=4(rem) answaer
• 9 years agoHelpfull: Yes(0) No(0)
• the remainder will be 4.hence option 1 is the ans.
• 9 years agoHelpfull: Yes(0) No(2)
• The Euler no. of 9 is =9 * 2/3 =6
  now 32/6 remainder=2 and 32/9 remainder is=5
  then (5^2)^32=25^32.
  again using Euler no. 32/6 leaves remainder 2.
  Then,
  (5^2)^2/9 =625/9 leaves remainder 4.
• 9 years agoHelpfull: Yes(0) No(0)