(9999999999)^3 what is the sum of the digits?

• 9^3 = > 7 + 2 + 9 = > 9*2 >18
  99^3 = > 9 + 7+ 0 + 2 + 9+9 =>9*4 = 36
  999^3 => 9 +9+7+0+0+2+9+9+9 => 9 * 6 = 54
  (999999...n) ^ 3 => 9 * 2n
  So in the given question number of 9 is 10 =>
  9*2*10 = 180.
• 8 years agoHelpfull: Yes(54) No(2)
• 9999999999^3 = (10^10-1)^3
  Where A=10^10 and B = 1
  Apply (A-B)^3
  A^3 -B^3 -3AB(À+B)
  
  = 10^30 - 1^3 -3x10^10x1(10000000001)=
  180
  Ans is 180
• 8 years agoHelpfull: Yes(7) No(3)
• (9999999999)^3 = 999999999700000000029999999999
  
  9+9+9+9+9+9+9+9+9+7+000000000+2+9+9+9+9+9+9+9+9+9+9=180
• 8 years agoHelpfull: Yes(3) No(12)
• sum of the digits of cube of 99^3 = (970299)36 in which 36 = 18 x no.of digits of the original number
  
  sum of the digits of cube of 999^3 = (997002999)54 in which 54 = 18 x no.of digits of the original number.
  
  similarly 9,999,999,999^3 = 18 x 10 = 180
  
• 8 years agoHelpfull: Yes(3) No(0)
• 180 is correct answer.its number system problems and its not a tough question.You can easily deal it.
• 8 years agoHelpfull: Yes(0) No(11)