Bunuel
If n is a positive integer, is n^3 - n divisible by 24?
(1) n is divisible by 6
(2) n = 8m + 4, where m is an integer
\(n^3 - n = n*(n^2 - 1) = (n-1)*n*(n+1)\) (i.e. Product of three consecutive Integers)Question: Is (n-1)*n*(n+1) divisible by 8?Hint: Product of three consecutive Integers is divisible by 8 when
1) Smallest of them (i.e. n-1 in this case) is Even i.e. when n is ODD OR
2) when n itself is divisible by 8
Statement 1: n is divisible by 6Since n my be 6 in which case (n-1)*n*(n+1) is
NOT divisible by 8 OR n my be 24 (multiple of 8 and 6 both) in which case (n-1)*n*(n+1) is
divisible by 8 hence
NOT SUFFICIENTStatement 2: n = 8m + 4, where m is an integeri.e. n = 4(2m+1), since 2m+1 is ODD for all Integer values of m therefore,
4(2m+1) will Neither be divisible by 8 nor will ever be ODD hence
Answer to the question is definitely NO hence
SUFFICIENTAnswer: Option B