If x is an integer, then x(x – 1)(x – k) must

[mami]

If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT

A -4
B -2
C -1
D 2
E 5

(Answer is B)

From answer explanation: "Note that the difference between k = -1 and k = 2 is 3. Every third consecutive integer would serve the same purpose in the product x(x - 1)(x - k): periodically serving as the multiple of three in the list of consecutive integers. Thus, k = -4 and k = 5 would also give us a product that is always divisible by three."

I eliminated k=-1 and k=2 from the answer choices since each of these would make x(x-1)(x-k) consecutive numbers, thus divisible by 3. I had trouble eliminating other answer choices. Please expand on the last sentence of the answer explanation above. Should I have looked for two answer choices that yield a distance/difference that is a multiple of 3?

[RonPurewal]

If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT

A -4
B -2
C -1
D 2
E 5

(Answer is B)

From answer explanation: "Note that the difference between k = -1 and k = 2 is 3. Every third consecutive integer would serve the same purpose in the product x(x - 1)(x - k): periodically serving as the multiple of three in the list of consecutive integers. Thus, k = -4 and k = 5 would also give us a product that is always divisible by three."

I eliminated k=-1 and k=2 from the answer choices since each of these would make x(x-1)(x-k) consecutive numbers, thus divisible by 3. I had trouble eliminating other answer choices. Please expand on the last sentence of the answer explanation above. Should I have looked for two answer choices that yield a distance/difference that is a multiple of 3?

note: in the following text, 'mo3' stands for 'multiple of three'. (i'm a lazy typist)

here's one approach:
first, realize that, if either x or (x - 1) is a mo3, then there is nothing to prove. so you need to assume that neither of those two numbers is a mo3, so that you actually have something to find.

fact: every third number is a mo3. so, if x and (x - 1) are not mo3's, then all the red numbers in the following sequence must be mo3's:

(...)
x - 5
x - 4
x - 3
x - 2
x - 1
x
x + 1
x + 2
x + 3
x + 4
x + 5
x + 6
x + 7
(...)

all of the choices are red except for b. (note the added twist from the fact that it's x minus k, so you have to reverse the signs of the answers to correlate them to the list above)

--

you are right on with the last comment, though. if you look at choices a, c, d, and e, they are all marching in lockstep, each one 3 greater than the previous one - until choice b comes along and spoils the party for everyone.