What Will Be the Remainder if 132^66^33 is divided by 5

• Just use -ve remainder theorem..
  132^66^33 can be written as (130+2)^66^33
  now 130 is completely divisible by 5 so its any power is also divisible....
  So remainder will be from 2^66^33
  This can be written as 4^33^33..
  Now this can be written as(5-1)^33^33/5=(-1)^33^33/5=(-1/5),its rem=4....
  
• 10 years agoHelpfull: Yes(48) No(5)
• unit dgit's of 132=2
  66^33 it's unit digits=6
  2^6=64
  64 is divided by 5 then remainder=4
• 10 years agoHelpfull: Yes(38) No(9)
• When u multiply 132*66 only find the last digit... It comes out to be *****2 and when u multiply this by 33 last digit will be 6 i.e *****6
  
  Now we know that when there is 5 or 0 at end, no. gets completely divided by 5 so when 6 is at unit place 1 will be the remainder
  
  Answer : 1
• 10 years agoHelpfull: Yes(8) No(8)
• ANSWER: 4 or 1.
  
  EXPLANATION: 66^33 will always have 6 in its ones place (just like 6 raised to the power of any number has 6 in its ones place). Now 132 raised to the power any number with 6 in its ones place(like 6,16,26,36,etc.) will have 4 or 6 in its one's place. Thus when divided by 5 132^66^33 will have 4 or 1 as remainder.
• 10 years agoHelpfull: Yes(5) No(1)
• using -ve remainder theorm
  2^66^33/5=(4)^33*33/5
  =-1/5=4
• 10 years agoHelpfull: Yes(3) No(0)
• 132^66^33 this whole number can be taken as 2^6^3 for findng the remainder (taking only unit placed digits in 132,66 and 33)
  units digit in 6^3 is 6
  units digit in 2^6 is 2
  2%5 is 2
  remainder is 2
• 10 years agoHelpfull: Yes(2) No(2)
• 66^33 it's unit digit=6^33=6
  last digit of 132=2
  2^6=64
  64 is divided by 5 then remainder is=4.
• 10 years agoHelpfull: Yes(2) No(0)
• answer is 1. As (even)^4k = last digit 6.(k=1,2,3,4,.....)
  so 6%5=1
• 10 years agoHelpfull: Yes(2) No(0)
• 132^66^33/5
  132=130+2
  so 2^66^33
  now 2^66 ends with 4
  so 4^33
  now 4/5 remainder is -1
  so -1^33/5
  final remainder -1
  so remainder will be 5-1=4
• 10 years agoHelpfull: Yes(2) No(0)
• Take the unit digit in 132
  2 has only 4 cycles and if 66 is divided by 5 we get remainder as 2 so in cycles of 2 we get 4.
  Next 4 has only a&6 as its cycles and when 33 is divided by 4 we get rem as 1
  Therefore 4/5=4 as the numerator is smaller than the denominator.
• 10 years agoHelpfull: Yes(1) No(1)
• 132/5=2
  2/5=2
  2^2/5=4
  2^3/5=3
  2^4/5=1.
  so cycle is 4.
  66/4=2
  132^66/5=4
  4/5=1
  4^2/5=1
  33/4=1
  4^33/5=4
  ans=4.
• 10 years agoHelpfull: Yes(1) No(0)
• 132^66^33
  First Calculate last two digits of 66^33 which come out to be 96
  so 132^***96/5=2^***96/5.... unit digit will come out to be 6 since 96 is divisible by 4 .... so ans has to be 6%5=1
• 10 years agoHelpfull: Yes(1) No(0)
• 32=130+2 i.e 130 is divisible by 5.Therefore will always leave remainder ZERO.
  Now, 2^66^33 will give us required remainder.
  Take 2^66 => 16^64 => On Dividing 16 by 5, Remainder=1
  Therefore, 1^64^33 is equals 1.
  Hence Remainder = 1
• 10 years agoHelpfull: Yes(0) No(3)
• 132^66^33
  first 66^33
  6 power anything is 6
  so 132^6
  2 has 4 possibilities so 6/4=1 remainder 2
  2^2 is 4 ans
  
• 10 years agoHelpfull: Yes(0) No(0)
• 1
  132^66 give unit digit as 4 as per power cycle
  (xxx4)^33 give unit digit as 6 as per power cycle of 4
  there for 1 is remainder
• 10 years agoHelpfull: Yes(0) No(4)
• 132^66^33--->(130+2)^66^33 Since 130 is completely divisible by 5, you only have to think about 2^66^33 again 4^33^33---> (5-1)^33^33 now 5 is completely divisible by 5 (-1)^33^33=-1 which means remainder is 5-1= 4 ans.
• 10 years agoHelpfull: Yes(0) No(0)
• 132^66^33/5=
  7^66^33/5=
  (7^2)^33^33/5=(-1)^33^33/5=-1/5=4
• 10 years agoHelpfull: Yes(0) No(0)