If n and m are positive integers, and 3 is not a factor of m, then m

Here is my approach
If n and m are positive integers, and 3 is not a factor of m, -> m is not divisible by 3-> answer wont be divisible by 3

A. (n−1)·n·(n+1) divisible by 3 because n-1, n, n+1 are 3 consecutive integers so always have 1 of them divisible by 3 then (n-1)n(n+1) divisible by 3
B. (n−3)·(n−1)·(n+1) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n-3)(n-1)(n+1) has the same remainder with (n-1)n(n+1) when divide by 3
Like choice A-> (n-3)(n-1)(n+1) divisible by 3
C. (n−2)·n·(n+2) divisible by 3
Similar to B, n-2 has the same remainder with n+1 when divide by 3
n+2 has the same remainder with n -1 when divide by 3
=> (n-2)n(n+2) divisible by 3
D. (n−1)·n·(n+2) => seem good. ex: n=5 then m is not divisible by 3 => CORRECT
E. (n−3)·(n+1)·(n+2) divisible by 3
n-3 has the same remainder with n when divide by 3
=> (n−3)·(n+1)·(n+2) has the same remainder with n(n+1)(n+2) when divide by 3
Have: n, n+1, n+2 are 3 consecutive integer then n(n+1)(n+2) is divisible by 3 => (n-3)(n+1)(n+2)


Hence IMO D

What happens when N = 3 ?

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The question is which maybe m. So when other options always divisible by 3 but only D may not.

Can anybody explain, what does this question mean exactly? how is m related to n, to find out m in relation to n? thank you.