CAT Exam

If n^3+100 is exactly divisible by n+10, where n is a positive integer.What is the maximum possible value of n ?

Read Solution (Total 1)

890
(n^3 + 100)/100 = n^2 - 10n + 100 - 900/( n+10)

so 900 must be divisible by n+10.

so max possible value of n= 890
12 years agoHelpfull: Yes(0) No(1)

CAT Other Question

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