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As we can see we have the product of 4 consecutive integers: (k – 1), k, (k+1), and (k + 2). Thus, the product will always be divisible by 4! = 24, which means that it will be divisible by 6 too.

Answer: A.

Of course we can simply plug-in numbers: since PS question cannot have more than 1 correct answer, then for any k the answer must be the same. For example, if k = 1, then the product becomes 0. 0 divided by 6 gives the remainder of 0 (0 is a multiple of every integer, except 0 itself).
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Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
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24 Jun 2017, 15:15

6

\((k + 2) (k^3 – k)\)

\((k + 2) (k (K^2 - 1))\)

\((k + 2) k (k + 1) (k - 1)\)

\((k - 1) k (k + 1) (k + 2)\)

As we can see that above numbers are 4 consecutive integers, and as we know \(4\) consecutive integers are multiples of \(4! = 4 * 3 * 2 * 1 = 24\)

Hence, as \(24\)is divisible by \(6\), equation will be commpletely divisible by \(6\) and hence reminder will be ZERO.

Hence, Answer is A

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If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
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Updated on: 13 Nov 2018, 08:21

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AbdurRakib wrote:

If k is a positive integer, what is the remainder when (k + 2)(k³ – k) is divided by 6 ?

A. 0 B. 1 C. 2 D. 3 E. 4

Alternatively, we can plug in any positive integer for k and see what happens.

Try k = 1 We get: (k + 2)(k³ – k) = (1 + 2)(1³ – 1) = (3)(0) = 0 When we divide 0 by 6 we get 0 with remainder 0 Answer: A

Just for "fun," let's test another k-value Try k = 2 We get: (k + 2)(k³ – k) = (2 + 2)(2³ – 2) = (4)(6) = 24 When we divide 24 by 6 we get 4 with remainder 0 Answer: still A

Try k = 3 We get: (k + 2)(k³ – k) = (3 + 2)(3³ – 3) = (5)(24) = 120 When we divide 120 by 6 we get 20 with remainder 0 Answer: still A

And so on...

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Originally posted by GMATPrepNow on 28 Jun 2017, 05:58.
Last edited by GMATPrepNow on 13 Nov 2018, 08:21, edited 1 time in total.

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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Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
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09 Jan 2018, 17:28

Hi All,

This question it perfect for TESTing VALUES (the approach that Abhishek009 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.

As we can see we have the product of 4 consecutive integers: (k – 1), k, (k+1), and (k + 2). Thus, the product will always be divisible by 4! = 24, which means that it will be divisible by 6 too.

Answer: A.

Of course we can simply plug-in numbers: since PS question cannot have more than 1 correct answer, then for any k the answer must be the same. For example, if k = 1, then the product becomes 0. 0 divided by 6 gives the remainder of 0 (0 is a multiple of every integer, except 0 itself).

ok so you say we hav a product of 4 consecutive integers (k – 1), k, (k+1), and (k + 2) but that one K is not in the form of (k-1) it is just single K :) why ?

for example i have this (k – 1), k, (k+1), and (k + 2) and will take 4 and plug in

so i get (4 – 1), 4, (4+1), and (4 + 2) --- > 3, 4, 5, 6

but if I take 0 for K i get ( 0 – 1), 0, (0+1), and (0+ 2) ---> i get 0 and then remain is 6 ?

As we can see we have the product of 4 consecutive integers: (k – 1), k, (k+1), and (k + 2). Thus, the product will always be divisible by 4! = 24, which means that it will be divisible by 6 too.

Answer: A.

Of course we can simply plug-in numbers: since PS question cannot have more than 1 correct answer, then for any k the answer must be the same. For example, if k = 1, then the product becomes 0. 0 divided by 6 gives the remainder of 0 (0 is a multiple of every integer, except 0 itself).

ok so you say we hav a product of 4 consecutive integers (k – 1), k, (k+1), and (k + 2) but that one K is not in the form of (k-1) it is just single K why ?

for example i have this (k – 1), k, (k+1), and (k + 2) and will take 4 and plug in

so i get (4 – 1), 4, (4+1), and (4 + 2) --- > 3, 4, 5, 6

but if I take 0 for K i get ( 0 – 1), 0, (0+1), and (0+ 2) ---> i get 0 and then remain is 6 ?

thank you

0 is divisible by every integer (except 0 itself): 0/integer = integer. So, 0 divided by 6 gives the remainder of 0.
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