30^72^87 divided by 11 gives remainder

• 72^87 = 10k + 8
  =>30^72^87 = 30(10k + 8) = 30^8mod11 = (-3)^8mod11 = 3^3mod11 [ 3^5 = 1mod11]
  = 27mod11 = 5mod11.
  so remainder is 5.
  
• 10 years agoHelpfull: Yes(35) No(21)
• 30^72^87=30^6264/11
  =(33-3)^6264/11
  using binomial theorem,
  the last term in the expansion is (-3)^6264/11
  =((-3)^2)^3132/11
  =9^3132/11
  =(11-2)^3132/11
  the last term in the expansion is
  =(2^(5*626)+2)/11=(32^626)*4/11
  ((33-1)^626)*4/11=4/11 remainder is 4
  
• 10 years agoHelpfull: Yes(34) No(39)
• 5 will be the correct answer
  7287 = 10k + 8
  => 3072^87 = 30(10k + 8) = 308mod11 = (-3)8mod11 = 33mod11 [ 35 = 1mod11]
  = 27mod11 = 5
• 10 years agoHelpfull: Yes(8) No(2)
• ans is 1
  
• 10 years agoHelpfull: Yes(7) No(18)
• Actually what i did wrong is 5*4*9=180
  Now, 180/11 will give remainder 4.
  So 4 is the correct answer.
  
• 10 years agoHelpfull: Yes(5) No(17)
• ans=5
  sollution:
  72^87=10k+a
  a=find the unit digit of this number=8
  means 72^87=10k+8
  30^(10K+8) where 30^10k remainder is 0
  then we take 30^8mod11=(-3)^8mod11=(3)^5*3^3mod11=1*3^3Mod11=5 ans
• 10 years agoHelpfull: Yes(5) No(0)
• Here the base is 30 and its exponent is 72^87.
  
  Find out the units digit of exponent i.e of 72^87 which will be 8.
  The problem is now reduced to 30^8.
  Expand the function in terms of factor of 11; (22 + 8)^8.
  In binomial expansion, all the term will be divisible by 11 except 8^8.
  Again expand it as 64^4 = (66 - 2)^4.
  Here also all the term will be divisible by 11 except (-2)^4 = 16.
  
  The final remainder will be 5
• 7 years agoHelpfull: Yes(5) No(1)
• For 3 multiples
  3 9 27 81 243 729
  
  Remainder when devided by 11:- 3 9 5 4 1 RESP
  
  For 2,
  2 4 8 16 32 64 128 256 512 1024 2048 4096
  
  
  Remainder when devided by 11:2 4 8 5 10 9 7 3 6 1
  
  For 5 multiples
  5 25 125 625 3125
  
  Remainder when devided by 11: 5 3 4 9 1
  
  5^5 WILL give reminder 1
  2^10 WILL GIVE REMINDER 1
  3^5 WILL GIVE REMINDER 1
  
  
  30^72^87/11=(2.3.5)72^87= 2^72^87.3^72^87.5^72^87
  
  2^6264.3^6264.5^6264/11=2^6260.2^4.3^6260.3^4.5^6260.5^4/11=2^4.3^4.5^4/11=2^4.15^4/11=2^4.4^4/11=2^12/11=4096/11
  ans=4
• 10 years agoHelpfull: Yes(4) No(3)
• 7 is the answer.
  
  It can b written as (2*3*5)^6264
  = 2^6264 *3^6264 * 5^6264
  Now when 2^4 divided by 11 the remainder is 5
  when 2^14 divided by 11 the remainder is 5
  when 2^24 divided by 11 the remainder is 5
  when 2^34 divided by 11 the remainder is 5
  and so on....
  
  Same u can do for 3 and 5
  
  so ((2^6264)/11) , remainder will be 5
  ((3^6264)/11) , remainder will be 4
  ((5^6264)/11) , remainder will be 9
  
  So total 5+4+9=18
  Now 18/9 will give remainder 7.
  
• 10 years agoHelpfull: Yes(3) No(28)
• 2multiples:2 4 8 16 32 64 128 256 512 1024 2048 4096
  Remainder when devided by 11:2 4 8 5 10 9 7 3 6 1 resp.
  
  So, 2^10 gives remainder as 1.
  
  
  30^72^87/11=(22+8)^6264/11 Remainder will be same as 8^6264/11
  
  8^6264/11= 2^(3*6264)=2^18792/11= 2^18790.2^2/11=2^2/11=4/11
  
  So,Remainder=4
• 10 years agoHelpfull: Yes(3) No(1)
• vikas..plzzz explain me in detail
• 10 years agoHelpfull: Yes(1) No(2)
• it is in d form a^m^n ... it is not in d form (a^m)^n = a^(m*n) .... how can you multiply 72 and 87 !!
• 10 years agoHelpfull: Yes(1) No(0)
• http://totalgadha.com/mod/forum/discuss.php?d=2594
  
  see it 4 this type of que
• 10 years agoHelpfull: Yes(1) No(1)
• 6. 30^72^87 divided by 11 gives remainder
• 10 years agoHelpfull: Yes(1) No(0)
• 7 should be the answer
  
• 9 years agoHelpfull: Yes(1) No(0)
• 3010 or 810when divided by 11 remainder is 1.
  The unit digit of 7287 is 8 (using cyclicity of unit digits) Click here
  So 7287 = 10K + 8
  30(10K+8)11=(3010)K.30811=1k.30811
  8811=22411=(25)4.2411=1611=5
• 9 years agoHelpfull: Yes(1) No(0)
• 
  This should take some thought. Let's start with 30 and work our way up. When you divide 30 by 11, you get a quotient of 2 and a remainder of 8.
  
  How about 302 ? It's 30 times 30. Each 30 divided by 11 gives a remainder of 8. So
  
  30⋅30=(2⋅11+8)(2⋅11+8)=∗⋅11+82
  
  where * is some number we don't care about. And 82=64 gives a remainder of 9 when divided by 11. Aha! The remainder for 302 is the same as the remainder for 82.
  
  Now we can go back to the earlier question. 87287 when divided by 11 gives the same as 88 when divided by 11, and according to our first table, that remainder was 5.
  
  Therefore, we have the answer to the original question. The remainder of 307287 when divided by 11 is 5.
  
  As I was not able to post the pic of table
  Please click here..
  http://linkshrink.net/7uQbvI
  I am sure you get solution so easily..
• 8 years agoHelpfull: Yes(1) No(0)
• 72^87 = 10k + 8
  => 3072^87 = 30(10k + 8) = 308mod11 = (-3)8mod11 = 33mod11 [ 35 = 1mod11]
  = 27mod11 = 5mod11.
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 1
  (30^72^78)/11
  (30^6264)/11
  ((3^5)^6259)/11
  (243/11)^6264 ---> R--> ((1)^6264)/11
  one raise to the power any number is one.
  
• 8 years agoHelpfull: Yes(0) No(0)
• ----- BY using fermets theorem
  a^(p-1)=1modp
  so , write the given qst as 30^6264
  p=11,a=30
  
  30^10=1mod11
  30^4 ((30^10)^626)=1^6260* 30^10 mod 11
  so
  30^10 mod 11=4
  so ans is 4.
  
  
• 8 years agoHelpfull: Yes(0) No(0)
• Fermat little theorem says, (ap−1)/p remainder is 1. ie., 3010 or 810when divided by 11 remainder is 1.
  The unit digit of 7287 is 8 (using cyclicity of unit digits) So 7287 = 10K + 8
  30(10K+8)/11=(3010)K.308/11=1k.308/11
  8^8/11=2^24/11=(25)4.24/11=16/11=5(we can write (2^5)^4 also as (2^10)^2/11=1) or
  then we take 30^8mod11=(-3)^8mod11=(3)^5*3^3mod11=1*3^3Mod11=5 ans. (i.e. either take 30-22=8 or 30-33=-3)
  
• 8 years agoHelpfull: Yes(0) No(0)
• ans=5
  because so many persons were solved the solution...
  then i also agree these type of answer
• 4 years agoHelpfull: Yes(0) No(0)