what is the remainder when 30^72^87 is divided by 11.
A) 5
B) 7
C) 1
D) None of these

• okay here is the general method of how to solve this type of problem
  
  eg. 23^56^78/11 or 63^54^78/6 or x^y^z/(any number) something like that
  
  so lets take the above problem
  
  30^72^87/11 .............. no need to go in detail just follow this method and all problem of this type will be solved.
  
  
  1. now always divide z or 87 or highest power with 4.
  
  we get 87/4 and take its remainder ,,, basically 87%4=3 and save it .
  
  2. now always divide 56 or y or middle power with 10 .
  
  we get 72/10 and take its remainder ,,,, basically 72%10 = 2
  
  3. now we get z=3 and y=2 ,, now solve this until you get a ( unit digit < 10 )
  
  2^3 or y^z (2^3 = 2 * 2 * 2 / 10 )
  
  4. now divide x or 30 by 11 and get remainder i.e. 30 % 11 = 8
  
  now put the solved y^z or 72^87 = 8 ,,,, here
  
  we get 8 ^ 8 % 11 = 5 -----> (8 * 8 * 8 * ............... 8 times % 11 )
  
  use this for all the problems of this type divide y by 10 and z by 4.
  
  and quickly get the answer .
• 8 years agoHelpfull: Yes(40) No(6)
• Ans is 5 option (a)
• 8 years agoHelpfull: Yes(16) No(7)
• let us take N=72^87
  eq is..30^N/11=8^N/11
  here cyclicity of 8 is 10
  then N/10=72^87/10=2^87/10
  here cyclicity of 2 is
  then 87%4=3(3rd cycle of 2)....i.e2^3%10=8(8th cycle of8)
  therefore final ans is 8^8%11=5
• 8 years agoHelpfull: Yes(14) No(3)
• Rem [30^72^87 / 11] = Rem [(-3)^72^87 / 11] = Rem [3^72^87 / 11]
  
  Now, we need to observe the pattern
  3^1 when divided by 11, leaves a remainder of 3
  3^2 when divided by 11, leaves a remainder of 9
  3^3 when divided by 11, leaves a remainder of 5
  3^4 when divided by 11, leaves a remainder of 4
  3^5 when divided by 11, leaves a remainder of 1
  And then the same cycle of 3, 9, 5, 4 and 1 will continue.
  If a number is of the format of 3^(5k + 1), it will leave a remainder of 3
  If a number is of the format of 3^(5k + 2), it will leave a remainder of 9
  If a number is of the format of 3^(5k + 3), it will leave a remainder of 5
  If a number is of the format of 3^(5k + 4), it will leave a remainder of 4
  If a number is of the format of 3^(5k), it will leave a remainder of 1
  
  The number given to us is 3^72^87
  Let us find out Rem[Power / Cyclicity] t0 find out if it 3^(5k + what?)
  
  Rem [72^87 / 5]
  = Rem [2^87 / 5]
  = Rem [2*4^43/5]
  = Rem [2*(-1) / 5]
  = -2
  = 3
  => The number is of the format 3^(5k + 3)
  => Rem [3^72^87 / 11] = 5
  
• 8 years agoHelpfull: Yes(8) No(1)
• i think it will be none of these...
  as 30/11 will give us reminder of 8
  now cyclicity of 8 when divided by 11 is 10 as
  8^1%11=8;
  8^2%11=9;
  8^3%11=(break)8^2*8^1=(as previous reminder)8*9=72%11=6;
  8^4%11=4;
  8^5%11=10;
  8^6%11=3;
  8^7%11=2;
  8^8%11=5;
  8^9%11=7;
  8^10%11=1;
  Now as the rule of powers 72*87=6264;
  we can break
  ((8^10)626)%11*(8^4)%11= (1)^626 * 4096%11=1*4
  so ANS will be 4....
  so ans is d...
  PLZ
  Notify me if any mistake....
• 8 years agoHelpfull: Yes(5) No(4)
• 30^72^87/11
  
  in this type of question u have get value 1^something then u can give the answer easily
  if u know cyclicity then u can get answer easily.
  first 30/11=8 reminder
  now this will become 8^72^87/11
  find the reminder of 8/11 if u get 1 then u came to answer
  8/11=3
  8^2=8*8/11=2
  8^3=5 ------(i)
  8^4=7
  8^5=1 here we get 1 ok
  now here cyclicity is 5
  so we have to dive 72^87/5= 2 as reminder then
  now cycle of 2 is 4 so 87/4=3 reminder now go to the ---(i)
  where u get the answer .... as 5.
  
  
• 8 years agoHelpfull: Yes(4) No(4)
• ok i have read all the solutions here
  some people were asking how to get 8^8/11 remainder?
  it is simple
  =(11-3)^8/11
  =(-3)^8/11
  =3^8/11
  =3^4*3^4/11
  =81*81/11
  =(77+4)(77+4)/11
  since 77 is divided by 11 so again it will become
  =4*4/11
  =16/11
  now you know remainder would be 5
  
• 8 years agoHelpfull: Yes(4) No(0)
• c) 1 is the ans.
• 8 years agoHelpfull: Yes(3) No(3)
• 72^87 can be written as 10k+8
  so 30^(10k+8)/11
  =30^(11-1)^k . 30^8/11
  now from the formula a^(p-1)/p gives a remainder 1 where p is a prime number.
  so 30^(11-1)^k gives remainder 1.
  now 30^8/11 gives remainder 5.( go through remainder theorem. you will find)
  so remainder is.... 5
• 8 years agoHelpfull: Yes(3) No(1)
• hi guys,i dont get how 8^8%11 gives the remainder as 5.i would be thankful if any1 spare a few time in explaining it.
• 8 years agoHelpfull: Yes(3) No(1)
• sry m not able to undrstand any solutn .. plz suggest me in brief :(
• 8 years agoHelpfull: Yes(3) No(2)
• polo bhai agar %4 lene me 0 ajaye to is question me kya krenge
  
  (50 ki power 56 ki power 52)%11
  
• 8 years agoHelpfull: Yes(2) No(0)
• Ans is 4 hence 4 is not there hence Ans is(D)None of this
• 8 years agoHelpfull: Yes(2) No(2)
• sol:
  30%11=8 72%11=6 87%11=10
  now 8*6*0=480%11=7
  so ans:7
• 8 years agoHelpfull: Yes(1) No(0)
• plz exaplain it
• 8 years agoHelpfull: Yes(0) No(0)
• Please explain the solution
• 8 years agoHelpfull: Yes(0) No(0)
• correct ans is 7
  devide each no individual like (30%11= -3) and (72%11= 6) and (87%11= -1)
  now multiply each remender and devide again get 7 remender
• 8 years agoHelpfull: Yes(0) No(3)
• please somebody explain me how to find 8^8%11=5.............. ill be very thankful
  
  
• 8 years agoHelpfull: Yes(0) No(0)
• sory
  i can not
• 8 years agoHelpfull: Yes(0) No(0)
• For those who wanted solution for 8^8%11=5
  8=2^3
  therefore we can write
  (2^3)^8 which is
  2^24
  now 2 repeats cycle after 4 i.e.
  2^1=2
  2^2=4
  2^3=8
  2^4=16
  2^5=32...repeats
  
  therefore 2^24 = 2^4=16
  now
  16%11=5
  Hence the answer
• 4 years agoHelpfull: Yes(0) No(0)