If \(k\) is a positive integer, what is the remainder when \((k + 2)(k^3 – k)\) is divided by 6?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

As a general rule PS (problem solving) remainder questions are relatively easy, and this problem is no exception. This is also an example of a theorem problem. There is something somewhat deep going on here, but an example is all we need to solve the problem. I’ll discuss this idea in a moment as it comes up most often in DS (data sufficiency) remainder problems which tend to be much more complex.

Anyway, first things first. Keep it simple: lets say \(k = 1\). Then \((k + 2)(k^3 – k) = (1 + 2)(1^3 – 1) = 0\). Zero is divisible by everything except 0, so the remainder must be 0.

**The correct answer is A.**

Now, the fancy stuff. Let’s factor that expression:

Factor a \(k\) out of \((k^2 – 1)\).

\((k + 2)(k^3 – k) = (k + 2)(k)(k^2 – 1)\)

Factor \(k^2 – 1\) using the difference of squares formula.

\((k + 2)(k^3 – k) = (k + 2)(k)(k + 1)(k – 1)\)

Rearrange:

\((k + 2)(k^3 – k) = (k – 1)(k)(k + 1)(k + 2)\)

That’s the product of four consecutive integers. A classic GMAT trick, though, like I said, much more common and more important in DS questions.

Now for a nice rule – *The product of n consecutive integers is divisible by n!*

Because we have four consecutive integers here, the product is divisible by \(4! = {4}\times{3}\times{2}\times{1} = 24\). Because 6 is a factor of 24, the product is also divisible by 6.