If n is a natural number, then (6n2 + 6n) is always divisible by? Pick:  A) 6 only  B) 6 and 12 both  C) 12 only  D) 6 or 12 only  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>B</result>
If n is a natural number, then (6n2 + 6n) is always divisible by?