What is the greatest positive integer n for which n3 + 102 is divisible by n + 10?

• By division we find that n^3 + 102 = (n + 10) (n^2 - 10n + 100) - 898. Thus, if n + 10 divides n^3 + 102, then it must also divide 898. Moreover, since n is maximized whenever n + 10 is, and since the largest divisor of 898 is 898, we must have n+10 = 898. Therefore, n = 888.
• 11 years agoHelpfull: Yes(30) No(0)
• We can apply the remainder theorem, which is when f(x) is divided by (x-a) the remainder will be f(a)
  
  So, when n ^ 3 is divided by n+10, which is (n-(-10), the remainder will be
  (-10) ^ 3 + 102 = -898
  
  So, the remainder is -898, if the number is 898 or a factor of it, the entire expression is divisible .
  
  So, the possible valus of (n+10) is 898
  So, the possible values of n is 888
  
  So, the maximum possible value of n is 260
  
  Answer is 888
• 10 years agoHelpfull: Yes(3) No(0)
• the greatest will be 4.
• 11 years agoHelpfull: Yes(2) No(4)
• answer is 887.....
  here after dividing (n^3+102)by n+10 we get
  quotient=n^2-10n+100 and remainder as =-898
  there fore n+10>-898
  by solving this...we get
  n+10
• 11 years agoHelpfull: Yes(2) No(1)
• (n^3+102)/n+10=(n^2-10n+100)-898/n+10
  so to divide 898 by n+10 maximum integer value of n should be 888
• 11 years agoHelpfull: Yes(2) No(1)
• n value is 2
  
• 11 years agoHelpfull: Yes(1) No(4)
• ans is 62.
  
• 11 years agoHelpfull: Yes(1) No(0)
• sorry it is 62
  
• 11 years agoHelpfull: Yes(1) No(3)
• Mr Praveen is absolutely correct.
• 11 years agoHelpfull: Yes(1) No(0)
• is that 3n or n^3..??
• 11 years agoHelpfull: Yes(0) No(0)
• mounica can u explain
• 11 years agoHelpfull: Yes(0) No(4)
• Which one is correct.. pls explain...
• 11 years agoHelpfull: Yes(0) No(0)