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OG If k is a positive integer

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by GMATGuruNY » Sat Aug 19, 2017 2:54 am
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.
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by [email protected] » Wed Aug 23, 2017 7:17 pm
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

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Re: OG If k is a positive integer

by Scott@TargetTestPrep » Fri Feb 14, 2020 1:44 pm
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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