This question is testing principles of CONSECUTIVE INTEGERS.
Rule: The product of any 3 consecutive integers will always be divisible by 3.
Reason: Every 3rd integer is a multiple of 3, so if there are 3 consecutive integers, one of them will be divisible by 3, and thus the entire product will be divisible by 3.
Products of consecutive integers are often written in the format n(n + 1)(n + 2), or (n - 3)(n - 2)(n - 1), etc. The structure x(x - 1)(x - k) will be evenly divisible by 3 whenever we've created a set of 3 consecutive integers.
So, the following would be sets of consecutive integers (rearranged in order):
(x - 2)(x - 1)(x) --> in this case, k = 2, so eliminate D.
(x - 1)(x)(x + 1) --> in this case, k = -1
These are the only 2 values that would create a direct set of 3 consecutives. However, we could also guarantee divisibility by 3 if one of the values was 3 more or 3 less than the consecutive value. Consider:
(4)(5)(6) is divisible by 3.
(4)(5)(6 + 3) --> (4)(5)(9) is also divisible by 3
(4)(5)(6 - 3) --> (4)(5)(3) is also divisible by 3
So, if x(x - 1)(x + 1) is divisible by 3, then x(x - 1)(x + 4) must also work. This would mean k = -4, so eliminate A.
And if x(x - 1)(x - 2) is divisible by 3, then x(x - 1)(x - 5) must also work. This would mean k = 5, so eliminate E.
The only answer choice remaining is B.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
