What is the largest positive integer n for which n3 + 100 is divisible by n+10?

• (n^3 + 100)=[(n^3 + 1000)-900]=[(n^3 + 10^3)-900]....[eqn(1)]
  
  Now,clearly (n+10) is a factor of (n^3 + 10^3),i.e,(n+10) clearly divides
  
  (n^3 + 10^3) for any +ve integer n..
  
  So,from [eqn(1)],we can conclude that (n+10) must divide 900,and then only
  
  (n^3 + 100) will be divisible by (n+10)..
  
  Now at this point,it is very easy to calculate the largest value of n for which
  
  (n+10) will divide 900.
  
  The largest value of n, (n+10)=900; =>n=890.
  
  So,largest positive integer n for which (n^3 + 100) is divisible by (n+10) is
  
  =890..
  
  Thanks to all..:-):-)
• 11 years agoHelpfull: Yes(17) No(1)
• By the division algorithm, n3+100 = (n+10)(n^2-10n+100)-900,
  so
  (n3 + 100)/
  (n + 10)
  = {(n+10)(n^2 - 10n + 100) -900}/(n + 10)
  = (n^2-10n+100)-(900/n+10)
  :
  If n3 + 100 is to be divisible by n + 10, then 900=(n + 10) must be an integer.
  Thus, the largest possible n for which this is true is n = 890.
• 11 years agoHelpfull: Yes(5) No(4)
• What is the largest positive integer n for which (n^3 + 100) is divisible by (n+10) ?
• 11 years agoHelpfull: Yes(1) No(1)
• Ans: n = 0
  The digit sum of 4444 is when remainder obtained 4444 divided by 9
  4444 = (45−1)44
  Each term is a multiple of 9 but the last term which is (−1)44 = 1
  So the digit sum of 4444 is 1.
  
  Now the divisibility rule for 3, 9, 27... is the sum of the digits should be divisible by 3, 9, 27 respectively. In each case the digit sum is either multiple of 3 or 9.
  So for any value of n > 1, the given expression is not divisible by 3n
  
• 9 years agoHelpfull: Yes(0) No(0)