If n is a natural number, then (6n2 + 6n) is always divisible by?  Pick one
A) 6 and 12 both	B) 6 only	C) 12 only	D) 6 or 12	E) None
(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.   6 and 12 both)
<result>A</result>
If n is a natural number, then (6n2 + 6n) is always divisible by?