What is the remainder when 2^35 is divided by 5?

• 2^1 % 5 = 2
  2^2 % 5 = 4
  2^3 % 5 => 8 % 5 = 3
  2^4 % 5 => 16 % 5= 1
  2^5 % 5 => 32 %5 = 2
  2^6 % 5 => 64 %5 = 4
  .
  .
  again it goes on like 2, 4, 3, 1,
  The cycle is 4 powers of 5 upto 5^4...
  
  35 ==> 4 * k + 3
  ie 4 * 8 + 3
  so the remainder is 3
  if u get 4*8 + 1 , remainder ll be 2
  if u get 4*8 + 2 , remainder ll be 4
  if u get 4*8 + 3 , remainder ll be 3
  if u get 4*8 + 4 , remainder ll be 1
  
  Remainder = 3
  
  Ans : 3
  
• 10 years agoHelpfull: Yes(50) No(0)
• note the remainders of each power is in a cyclic manner...
  for power 1 , rem = 2
  for power 2 , rem = 4
  for power 3 , rem = 3
  for power 4 , rem = 1
  
  for power 5, again rem = 2
  for power 6, rem = 4
  for power 7, rem = 3
  for power 8, rem = 1
  
  for power 9, again rem = 2
  nd it goes on like 4, 3, 1
  
  so upto 2^4, 1 cycle get over ==> {2^1=2, 2^2=4, 2^3=3, 2^4=1]
  so upto 2^8, 2 cycles get over [4*2] ==> {2^5=2, 2^6=4, 2^7=3, 2^8=1]
  so upto 2^12, 3 cycles get over [4*3] ==> {2^9=2, 2^10=4, 2^11=3, 2^12=1]
  so upto 2^16, 4 cycles get over [4*4] ==> {2^13=2, 2^14=4, 2^15=3, 2^16=1]
  
  so if u want to find the remainder of 2^13/5... wat u can do is
  cycle is upto 4th power
  so 13 ==> 4k+1, where k is any integer that denotes the completed cycles
  13 = 4*3 +1
  here k=3... which means 3 cycles got over...
  [ie] upto 2^12 , 3 cycles got over
  so 2^13 is again 2
  
  4k+1 = 2
  4k+2 = 4
  4k+3 = 3
  4k+4 = 1
  
  so here the power is 35
  the cycle is upto 4th power... so u have to multiply 4 with any integer to get the power value "35"
  so 35 can be written as ==> 4k+3
  35 --> 4*8 + 3
  so upto 2^32 ==> 8 cycles got over
  for 2^33, rem = 2
  for 2^34, rem = 4
  for 2^35, rem = 3
  
  hence rem = 3
  
  
• 10 years agoHelpfull: Yes(10) No(0)
• 2^35 / 5
  
  = 4^17 *2 / 5
  
  = (5-1)^17 *2 / 5
  
  = (-1)^17 *2 / 5
  
  = -2 / 5
  
  => remainder = -2+5 = 3
• 10 years agoHelpfull: Yes(8) No(0)
• Ans :3
  unit digit in 2^10 is 4
  so as for 2^30 --> 4...
  2^31 --> 4*2=8
  2^32 --> 8*2=6
  2^33 --> 6^2=2
  2^34 --> 2*2=4
  2^35 --> 4*2=8
  
  8/5 rem is 3
  
• 10 years agoHelpfull: Yes(6) No(0)
• 2^35= (2^5)^7 ..........................................(i)
  
  where 2^5=32 : unit place is 2
  from equation (i) (2)^7=128
  in 128 unit place is 8
  so 8/5 gives remainder 3 :)
  
• 9 years agoHelpfull: Yes(4) No(0)
• guys.. remember it has to be short cos derz just 1 min for 1 question...
  the way i solved in d test was dis way..
  
  2^1= 2
  2^2= 4
  2^3= 8
  2^4=16
  after dis last digit starts repeating itself 2,4,6,8..again (ex :2^5=32)
  
  its in a set of 4 terms...
  
  the power here is 35. the number less than 35 and divided by 4 is 32.
  
  so 2^32 last digit will be "6" itself.
  so 35-32=3. three more terms remaining now..
  
  its simple do it.... 2*2*2=8
  remainder wen 8 divided by 5 is 3.
  
  soo answer is 3.
• 10 years agoHelpfull: Yes(2) No(1)
• 2^35/5
  =32^7/5
  =(5*6+2)^7/5
  =>2^7/5
  =128/5=>remainder=3
• 10 years agoHelpfull: Yes(1) No(0)
• 3
  2^10*2^10*2^10*2^5
  4*4*4*2
• 7 years agoHelpfull: Yes(1) No(0)
• 2^5=32*
  2^3
  =====16/5
  rem====1
• 10 years agoHelpfull: Yes(0) No(0)
• 
  "The cycle is 4 powers of 5 upto 5^4..."
  i can't understand this one
• 10 years agoHelpfull: Yes(0) No(0)
• a pattern 2,4,8,6 and again 2,4,8,6
  2^35 = 2
• 9 years agoHelpfull: Yes(0) No(1)
• it can be easily solved by the unit digit terms..
  for 2.. the unit digits are :
  
  2^1=2 2^2=4 2^3=8 2^4=6
  2^5=2 2^6=4 2^7=8 and the cycle goes on....
  
  here the given power is 35.. 35/4=8.. which is 32
  remaining 3..
  
  similarly it can be applied for all numbers.
• 9 years agoHelpfull: Yes(0) No(0)
• 3 is the remainder
• 9 years agoHelpfull: Yes(0) No(0)
• remainder will be 3.....because cyclicity of 2 is 4 and so 2^35 will end with 8...... so remainder will be 3
• 9 years agoHelpfull: Yes(0) No(0)
• last digit in power of 2-- 2,4,8,6.So 35%4=3.Means last digit is 8.
  Now 8%5=3.answer
• 8 years agoHelpfull: Yes(0) No(0)
• in unit place of 2^35 is 8 so remainder is 3
• 8 years agoHelpfull: Yes(0) No(0)
• 2^1=2; 2^2=4; 2^3=8; 2^4=16; same unit digit will continue.
  so 2^35 = 35%4= 3
  now 2^3=8
  so 8%5=3 ans.
• 7 years agoHelpfull: Yes(0) No(0)