find remainder when (10^3+9^3)^1000 divided by 12^3

• (10^3+9^3)^1000 / 12^3
  
  = 1729^1000 / 1728
  
  => (1728*1+1)^1000 / 1728
  => 1^1000 / 1728
  => rem = 1
• 9 years agoHelpfull: Yes(100) No(4)
• Ans:1
  10^3 UNIT DIGIT IS 0
  9^3 UNIT DIGIT IS 9
  (0+9)=9
  
  (9)^1OOO=9^n unit digit is 1 9 1 9 1 9 1 9
  
  
  so (9)^1OOO unit digit is=9
  
  12^3 unit digit is =8
  
  (10^3+9^3)^1000 divided by 12^3=9/8=remainder 1
  
• 9 years agoHelpfull: Yes(41) No(12)
• @sahitya
  (1728*1+1)^1000/1728
  (1728+1)
  you take last digit in this term (1728+1)
  so (1)^1000=1
  1/1728=reminder is 1
  
  using google calc
  
  https://www.google.co.in/?gfe_rd=cr&ei=t1X7U4KlBavM8gftyoDoBQ&gws_rd=ssl#q=1+mod+1278
  
• 9 years agoHelpfull: Yes(5) No(0)
• 10^3+9^3 = 1729
  
  12^3=1728
  
  remainder will be 1^1000 = 1
  
• 9 years agoHelpfull: Yes(4) No(0)
• @vignesh
  (10^3)=1000
  (9^3)=729
  1000+729=1729
  we can write the 1729 in following from
  (1728*1+1)^1000/1728
  Take (+1)^1000 only
  So (1)^1000=1
  1/1728=reminder is 1
• 9 years agoHelpfull: Yes(3) No(0)
• =(1729)^1000/1728
  =(1728+1)^1000/1728=1^1000/1728=1
  
• 9 years agoHelpfull: Yes(2) No(0)
• Unit digit of 10^3 is 0 while that of 9^3 is 9...hence unit digit of((10^3+9^3) is 9...
  Now 9^1000gives a unit digit 1 .hence the remainder is 1
• 9 years agoHelpfull: Yes(1) No(0)
• 10^3+9^3=1729
  
  1729^1000/1728 { 12^3=1728}
  (1728+1)^1000/1728
  
  1^1000/1728
  
  1 is the req rem.Ans
• 9 years agoHelpfull: Yes(1) No(0)
• 10^3+9^3=1729
  12^3=1728
  (1729)^1000/1728=(1728+1)^1000/1728
  By remainders theroem:
  rem=1
• 9 years agoHelpfull: Yes(1) No(0)
• it can be written as
  (((10*10*10)/(12*12*12))+((9*9*9)/(12*12*12)))^1000
  now
  10/12=remainder is -2
  9/12=remainder is -3
  then,
  we can write
  ((-2*-2*-2)+(-3*-3*-3))^1000
  =(-35)^1000
  which means 1^1000
  so the remainder is 1
• 9 years agoHelpfull: Yes(1) No(0)
• (10^3+9^3)^1000=1729^1000
  12^3=1728
  when 1729%1728=1
  1^1000=1
  and 1%1728=1ans
  
  
  => (1728*1+1)^1000 / 1728
  => 1^1000 / 1728
  => rem = 1
• 9 years agoHelpfull: Yes(0) No(0)
• @rakesh: can u please explain the last two steps, i mean
  1^1000/1728
  rem=1 ??? how?
• 9 years agoHelpfull: Yes(0) No(0)
• @ karthikeyan i have doubt in this step (1728*1+1)^1000/1728
  (1728+1)
  you take last digit in this term (1728+1)
  so (1)^1000=1
• 9 years agoHelpfull: Yes(0) No(0)
• 1729^1000
  For every 2 powers last digit will become 9
  So 1729^1000 have 9 as remainder
  For 12^3, last digit is 8
  When 9 is divided by 8, the remainder is 1.
  
  ANS: 1
• 9 years agoHelpfull: Yes(0) No(0)
• (10^3+9^3)^1000 / 12^3
  
  = >{10^3+9^3)^10 / 12}^3
  
  => [ {1729}^10/12 ]^3
  => [ {1728 + 1}^10/12 ]^3
  => [ {1}^10/12 ]^ 3
  => 1 is the req rem. Ans.
  
• 9 years agoHelpfull: Yes(0) No(1)