Divisibility problem

My approach:
to be sure that the product of 3 numbers is divisible by 3 these 3 numbers must be consequtive. One of them will be a multiply of 3.
A. n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1) - it is a multiply of 3

Draw it on the number line and you'll see

A
Correct?

ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"

4meonly is right. It's tricky problem and would take some time to solve. If I get problem like that, I'd come up with some generic number larger than 6, say 10 and do calculations. It'd be faster in some cases

gmat009 wrote:ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"

n * (n + 1) * (n - 4) = n(n+1)(n-1-3)=n(n+1)(n-1) - (n+1)*3n

PussInBoots wrote:

gmat009 wrote:ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"

n * (n + 1) * (n - 4) = n(n+1)(n-1-3)=n(n+1)(n-1) - (n+1)*3n

Got it..
Thanks....

Pick 7 and 8 for n and you can eleiminate everyhting but A.

Like one of the other user mentioned the process of elimination will work here.