Gurpinder 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)

OA: A

please explain how you would handle such a problem

Plug values into each answer choice. Eliminate any answer choice that doesn't give us a factor that is a multiple of 3.

Plug in n=7:

A) 7*8*3. Since 3 is multiple of 3, hold onto A.
B) 7*9*6. Since 9 and 6 are each a multiple of 3, hold onto B.
C) 7*10*2. Since neither 7 nor 10 nor 2 is a multiple of 3, eliminate C.
D) 7*11*5. Since neither 7 nor 11 nor 5 is a multiple of 3, eliminate D.
E) 7*12*1. Since 12 is a multiple of 3, hold onto E.

Now plug n=8 into A, B and E only:

A) 8*9*4. Since 9 is a multiple of 3, hold onto A.
B) 8*10*7. Since neither 8 nor 10 nor 7 is a multiple of 3, eliminate B.
E) 8*13*2. Since neither 8 nor 13 nor 2 is a multiple of 3, eliminate E.

The correct answer is A.

But how can one prove in general?

GMATGuruNY wrote: Plug values into each answer choice. Eliminate any answer choice that doesn't give us a factor that is a multiple of 3.

Plug in n=7:

A) 7*8*3. Since 3 is multiple of 3, hold onto A.
B) 7*9*6. Since 9 and 6 are each a multiple of 3, hold onto B.
C) 7*10*2. Since neither 7 nor 10 nor 2 is a multiple of 3, eliminate C.
D) 7*11*5. Since neither 7 nor 11 nor 5 is a multiple of 3, eliminate D.
E) 7*12*1. Since 12 is a multiple of 3, hold onto E.

Now plug n=8 into A, B and E only:

A) 8*9*4. Since 9 is a multiple of 3, hold onto A.
B) 8*10*7. Since neither 8 nor 10 nor 7 is a multiple of 3, eliminate B.
E) 8*13*2. Since neither 8 nor 13 nor 2 is a multiple of 3, eliminate E.

The correct answer is A.

this is the only way to solve this type of problem.... can you explain me clearly.....