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

• We can use cyclic remainder theorem
  2^1 % 5 = 2
  2^2 % 5 = 4
  2^3 % 5 = 3
  2^4 % 5 = 1
  
  Again
  2^5 % 5 = 2
  2^6 % 5 = 4
  So the 1st cycle is upto the power of "4"...
  So the remainder of 2^35 % 5 ll be (4k + n)th value...
  35 => 4*8+3 [k=8, n=3]
  If n=1, rem=2
  If n=2, rem=4
  If n=3, rem=3
  If n=4, rem=1
  Hence the remainder is 3
  
  Ans : 3
• 10 years agoHelpfull: Yes(50) No(0)
• 2^35 / 5
  => (2^4)^8 * 2^3 / 5
  => (5*3+1)^8 * (5+3) / 5
  => (1*3)/5
  => 3/5
  => remainder = 3
• 10 years agoHelpfull: Yes(18) No(0)
• 2^4 / 5 is always have reminder 1
  so convert in this form like
  
  (((2^4)^8)*(2^3))/5
  ((2^4)^8)/5 remainder is 1
  1*(2^3)/5
  8/5=3
  answer is 3
• 10 years agoHelpfull: Yes(8) No(0)
• always follow this method for finding the remainder.
  eg: 2^35/5
  
  2^1=2 2/5=2
  2^2=4 4/5=4
  2^3=8 8/5=3
  2^4=16 16/5=1
  2^5=32 32/5=2
  
  the sequence repeats like 243124312431....
  
  so 2 4 3 1 2 4 3 1 2 4
  3 1 2 4 3 1 2 4 3 1
  
  1st series is for 1-10 2nd series is for 11-20
  then again 1st series for 21-30..
  
  now 2^35 second series 5th element..
  Its 3.
  
• 10 years agoHelpfull: Yes(6) No(1)
• remainder is 3. since 2^1=2
  2^2=4
  2^3=8
  2^4=16
  2^5=32
  2^6=64... so on
  so the last digit with increasing power repeats as 2,4,8,6 and again the same. Dividing 35 by 4 gives 3 as remainder so the last digit have been 8,any digit ending with 8 when divided by 5 gives remainder as 3.ans
  
• 10 years agoHelpfull: Yes(5) No(0)
• 2^4=unit digit is 6
  2^35=[2^4]^8*2^3= 6*8=unit digit is 8
  so rem is 8/5=3
  reminder=3
• 10 years agoHelpfull: Yes(3) No(0)
• ans: 3
  2^1=r(2)
  2^2=r(4)
  2^3=r(3)
  2^4=r(1)
  2^5=r(2).....
  2^35=r(3)
• 10 years agoHelpfull: Yes(2) No(0)
• last digit of 2^35=8 so rem=3
• 10 years agoHelpfull: Yes(2) No(1)
• Ans:3
  
  (2^10*2^10*2^10)*2^5
  (1024)^3*2^5
  Take unit digit in 1024
  4*2^5=4*32=128
  
  128/5=remainder is 3
• 10 years agoHelpfull: Yes(2) No(0)
• 2^4=16
  16 mod 5=1
  2^35=(2^4)^8*2^3
  (1)^8*3= 3
  ans =3
• 10 years agoHelpfull: Yes(2) No(0)
• as per the rule cyclic power..
  
  35%4=3 as remainder
  which means 2^3=8 and 8%5=3 as remainder.
• 9 years agoHelpfull: Yes(2) No(0)
• 2^5 reminder=2
  2^10 reminder=4
  2^15 reminder=3
  2^20 reminder=1
  2^25 reminder=2
  2^30 reminder=4
  2^35 reminder=3
  so the answer is 3
• 10 years agoHelpfull: Yes(1) No(0)
• 35/4=3
  Now take 2^3=8
  Next Divide 8 by 5
  8/5=3
  Therefore Remainder is 3
• 9 years agoHelpfull: Yes(1) No(0)
• The unit digit of
  2^1=2
  2^2=4
  2^3=8
  2^4=6
  2^5=2
  2^6=4
  2^7=8
  2^8=6
  this unit digit value repeats for every 4 powers-power cycle
  ((2^4)^8)*(2^3)
  take only the unit digit values,we get
  (6)*8=48(refer above)
  divide 48 by 5
  48%5=3
  Ans:3
• 10 years agoHelpfull: Yes(0) No(0)
• rem: 2^1/5=2 2^2/5=4 2^3/5=3 2^4/1 ....
  35/4=rem(3)
  ans:3
• 10 years agoHelpfull: Yes(0) No(0)
• power cycles and remainder cycle of 2 is:
  2 1
  4 2
  8 3
  6 0
  to get the unit digit 35/4(bcoz no of power cycle is 4) the remainder is 3.power cycle 3 is equal to 8.so 8/5 remainder is 3. ans is 3
  
• 10 years agoHelpfull: Yes(0) No(0)
• 2^1%5=2
  2^2%5=4
  2^3%5=3
  2^4%5=1
  so, 2^35=((2^4)^8)*(2^3)
  =((1)^8)*(3)
  =1*3
  =3
  
• 10 years agoHelpfull: Yes(0) No(0)
• 2^35=(2^4)^8*2^3
  16*8/5=1*3
  =3
• 9 years agoHelpfull: Yes(0) No(0)
• rem 3.... cyclic of 2 is 4........ 35/4 gives 3rd cycle of 2 which is 8... dividing 8 by 5 gives 3 remainder
• 9 years agoHelpfull: Yes(0) No(0)
• divide 35 by 4 so that remainder will be 8
  take the remainder 8 and divide it by 5 then we will get the answer 3
• 6 years agoHelpfull: Yes(0) No(0)