gregspirited wrote:
If n is an integer greater than 6, which of the following must be divisible by 3?
A. \(n (n+1) (n-4)\)
B. \(n (n+2) (n-1)\)
C. \(n (n+3) (n-5)\)
D. \(n (n+4) (n-2)\)
E. \(n (n+5) (n-6)\)
all the above posts are brilliant but knowing this is not the concern. remembering to approach such question this way is the concern. if you're not the Math whiz such as Bunuel, you will, as I did, plug in values which is a BIG No-No.
product of three consecutive number is ALWAYS divisible by 3. but hey, none of these values aren't consecutive.... well, one of the sly one is
:
A. \(n (n+1) (n-4)\)
we can also write n-4= n+2-6
so, A becomes: \(n (n+1) (n+2-6)\)
now:
\(n (n+1) (n+2-6)\) << look at the highlighted part, it's product of three consecutive number.
and the remaining "-6" is also=\( 3*-2\) (meaning, multiple of 3) so this MUST be divisible by 3.
again, the challenge is not to KNOW or understand this. it's easy. the challenge is when in a middle of the test you will see such a question, can you approach it this way?