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 A

AbeNeedsAnswers wrote: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 We can test ANY POSITIVE INTEGER for k. Let k=1, with the result that (k+2)(k³-k) = (1+2)(1³-1) = (3)(0) = 0. In this case: [(k+2)(k³-k)]/6 = 0/6 = 0 R0. The correct answer is A.

Hi AbeNeedsAnswers, This question it perfect for TESTing VALUES (the approach that Mitch used). The question is also based on a subtle Number Property rule about consecutive integers. (K+2)(K^3 - K) can be rewritten as.... (K+2)(K)(K^2 - 1) = (K+2)(K)(K+1)(K-1) When multiplying numbers, the 'order' doesn't matter, so you can re-order those 4 'pieces' as... (K-1)(K)(K+1)(K+2) We're told that K is a positive integer, so what you have here is the product of 4 consecutive integers (they'll either all be positive or it's be 0-1-2-3) and EVERY 4 consecutive integers will include at least two multiples of 2 and a multiple of 3. By extension, this product will be some multiple of (2)(3) = 6, so when you divide the product by 6 you'll get a remainder of 0. Final Answer: A GMAT assassins aren't born, they're made, Rich

AbeNeedsAnswers wrote: ↑Fri Aug 18, 2017 5:36 pm 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 A Let’s simplify the given expression: (k + 2)(k^3 – k) = (k + 2)[k(k^2 - 1)] = (k + 2)(k)(k + 1)(k - 1) Reordering the factors in the expression, we have: (k - 1)(k)(k + 1)(k + 2), which is a product of 4 consecutive integers. Since the product of n consecutive integers is always divisible by n!, the product of 4 consecutive integers is always divisible by 4! = 24 and hence by 6. Thus, the remainder when (k + 2)(k^3 – k) is divided by 6 is 0. Answer: A

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OG If k is a positive integer

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