What is the remainder when 135^77 is divided by 7 ? I am getting ANS : 6.

• 135^1 % 7 = 2
  135^2 % 7 = (2*2)%7 = 4 [since 135^2 can be written as 135^1*135^1... So 2*2]
  135^3 % 7 = (4*2)%7 = 1
  
  135^4 % 7 = (1*2)%7 = 2
  135^5 % 7 = (2*2)%7 = 4
  135^6 % 7 = (4*2)%7 = 1
  
  Here the cycle is upto the power of 3
  2nd cycle again started with "2" nd it goes on like 4,1
  So the remainder ll be "3n+k" [since 1st cycle completed with the power of "3"]
  77 => 3(25)+2
  So upto 75(ie 3(25)), 25 cycles got over
  Next cycle again begin with "2"
  135^76 % 7 = 2
  135^77 % 7 = 4
  Hence reminder = 4
  
  Ans : 4
  
• 10 years agoHelpfull: Yes(83) No(6)
• (15^77 * 9^77)/7;
  (15^77)/7 * (9^77)/7;
  1 * 2^77/7;
  ((2^3)^25 * 2^2)/7;
  1 * 4 = 4
• 10 years agoHelpfull: Yes(18) No(3)
• 135^77/7
  now 135/7 gives reminder as 2
  2^77/7, we can write 2^5.2^72/7
  2^5(8^24)/7
  then [2^5*(7+1)^24]/7
  2^5/7
  remainder is 4
  
• 10 years agoHelpfull: Yes(8) No(1)
• general formula is x^(n-1)%n=1
  135^6%7=1 so 135^72%7=1
  remaining term is 135^5%7
  (15^5%7)*(9^5%7)=!
  15%7=1 and 9%7=2
  so (1^5%7)*(2^5%7)
  32%7=4
  ans is 4
• 10 years agoHelpfull: Yes(4) No(0)
• simple steps are..
  135 can be written as 15*9
  
  so (15*9)^77/7
  now dividing 15 and 9 with 7 we get remainder as 1 and 2 respectively
  so (1*2)^77/7
  
  or (2)^77/7
  now (2)^77 can be written as ((2)^3)^25 and rest will be 2^2
  i.e ((8)^25)*(2)^2/7
  or (7+1)^25*4/7
  so whe we divide only 1 will remain that will give 1^25*4/7 that will give 1*4/7
  so 4 will be ans..
• 10 years agoHelpfull: Yes(4) No(0)
• ans is 2
  135^77/7
  135 and 77 are co prime to each to other...so we can apply euler method to find the remainder
  E(7)=6
  77/6 the remainder is 1
  thus 135^1/7 is 2....
  answer is 2
• 10 years agoHelpfull: Yes(2) No(4)
• 135^77/7
  =(19*7+2)^77/7
  =2^77/7
  =(2^3)^25*2^2/7
  =4
• 10 years agoHelpfull: Yes(2) No(0)
• There is a shortcut method to find these type of problems
  
  Whenever a number ends with 5 chk it's ten's digit :
  Case 1:If it's a odd digit then last two digits of the number(here 135) raised to any exponent (here 77) will be 25.
  Case 2:if it's an even digit then last two digits of the number will either be 25 or 75.
  Here in this question we got 137^77=xxx25.
  Divide xxxx25 by 7 and u will get remainder as 4.
• 10 years agoHelpfull: Yes(2) No(0)
• 135/7 remainder=2
  2^77/7
  =((2^3)*2^74)/7
  =(2^3)^24*(2^4)[as 8/7 gives remainder 1 so 8^24 gives 1.so the rest part
  gives us 4/7]
  4/7
  =4
  remainder=4
• 10 years agoHelpfull: Yes(1) No(0)
• 135^77%7=(2^77)%7=((8^25)*4)%7=(1*4)%7=4
• 10 years agoHelpfull: Yes(1) No(1)
• (135)^77 =(7*19 +2)^77=2^77/7
  (2^3)^25 * 2^2/7=1*4=4
  hence rem is 4
• 10 years agoHelpfull: Yes(0) No(0)
• firstly divide 135 by 7 and take out the remainder. here remainder is 2.
  2^77/77=(2^3)^25.2^2 . now when 8 is divided by 7 we get 1 as remainder. thus remaining term 2^2 when divided by 7 we get 4 as remainder.
  
• 10 years agoHelpfull: Yes(0) No(0)
• (15^77 * 9^77)/7;
  (15^77)/7 * (9^77)/7;
  1 * 2^77/7;
  ((2^3)^25 * 2^2)/7;
  1 * 4 = 4
• 10 years agoHelpfull: Yes(0) No(0)
• according to formula : f(x)/x-1=f(1)
  135^77 divided by 7 can be written as: (2^3)^25 * (2^2)/2^3 - 1=f(1)*2^2=4,hence the answer is 4
  
• 10 years agoHelpfull: Yes(0) No(0)
• answer is 4
• 10 years agoHelpfull: Yes(0) No(0)
• take the last digit in 135 i.e 5 and using power cycles find out last digit i.e 5^7 is 5 now divide this last digit by 7 remainder is 2 so 2 is the answer
• 10 years agoHelpfull: Yes(0) No(2)
• ans.3
  
  (133+2)^77 %7=2^77%7=(-1)^25 *4%7=-4%7=3
• 10 years agoHelpfull: Yes(0) No(3)
• (135^77)/7)=>(2^77)/7=>[{(2^3)^25}*4]/7=>(1*4)/7=>rem=4
• 10 years agoHelpfull: Yes(0) No(0)
• 135^1%7=2
  Then according to the remainder theorem
  2^77%7=(2^7)^11%7=2^7%7=128%7=2
  then
  2^11 %7= (2^5 * 2^6)%7=(32%7)*(64%7)=4*1=4
• 10 years agoHelpfull: Yes(0) No(1)
• why not 6?
• 10 years agoHelpfull: Yes(0) No(0)
• i also getting 6
  
• 10 years agoHelpfull: Yes(0) No(0)
• there is a rule if p is a prime no then a^(p-1)/p gives remainder 1
  so first we got 77/6=(6*12)+5
  so it will become (135^6)^12 * 135^5
  the term before the * sign becomes 1
  no it remains 135^4
  there is another rule if u have to find the remainder in the case (a*b*c*d)/N
  then the answer will be multiplication of remainder in each separate division
  that means (a/N)*(b/N)*(c/N)....
  so if we divide 135 by 7 we get 2 as remainder. as their is 135^5 we have to multiply 2 by five times and we will get 32
  again divide 32 by 7 and the final remainder is 4
  ans: 4
• 8 years agoHelpfull: Yes(0) No(0)