If k is a positive integer, what is the remainder when (k + 2)(k^3 – k

There is a rule saying that: The product of n consecutive integers is always divisible by n!.

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

As we can see 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).
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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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\((k + 2)(k^{3}\) – k) \(= (k + 2) (k (k^2-1))\)

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

or could be written as

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

Its product of 4 consecutive numbers. Product of 3 consecutive numbers is always divisible by 6, Hence given number is divisible by 6.

Therefore remainder will be 0. Answer (A).

0. As k can be shown as product of three integers. Always divisible by 6

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Did you mean that k is a product of three *consecutive* integers? Then you are correct, always divisible by 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

One Kudos if you Like
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\((k + 2)(k^3 – k)\)
= \((k + 2)k(k^2 – 1)\)
= \((k + 2)k(k – 1)(k + 1)\)

Plug in numbers and check -

Let k = 2

\((k + 2)k(k – 1)(k + 1)\) = \(4*2*1*3\) = \(24\) ( Divided by 6 , remainder will be 0)

Let k = 3

\((k + 2)k(k – 1)(k + 1)\) = \(5*3*2*4\) = \(120\) ( Divided by 6 , remainder will be 0)

Thus, the answer will be (A) 0
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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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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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi Bunuel

can you please explain one thing. pleeeese

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

now, we know, the product of 3 consecutive numbers is always divisible by 6.
so, (k+2)(k^3–k) is divisible by 6.

the remainder when (k + 2)(k^3 – k) is divided by 6 = 0

correct answer A

Since the questions asks What is the remainder => the value will be same whatever value of " K " is.

If we put K = 1

( 1 + 2 ) ( 1^3 - 1 ) / 6 = 0/6 = 0

Answer A
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I did not notice that the numerator could be broken down into 4 consecutive integers. Is this alternative method OK and using sound logic?

\(\frac{(k + 2)(k^3 – k)}{6}\) = \(\frac{k^4 +2k^3 - k^2 - 2k}{6}\)

Since there are no common factors between the numerator and denominator, and there is at least one power of \(k\) in each numerator term, there will be a remainder of \(0\).

....................
If this is not OK logic, how was I supposed to be keyed in to notice that the numerator was a consecutive integer product? I did not think about that at all until I saw every solution on this thread using that method.

Thank you!

Answer: Option A

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Since K can be any positive integer, the remainder should be the same for any value of k. So we can use plug in method to solve this question.

If k =2, (k + 2)(\(k^3\) – k) = (2+2) ( \(2^3\) -2) = 4*6 =24

The remainder will be 0, when 24 is divided by 6.

Correct answer is A.

Thanks,
Clifin J Francis,
GMAT SME
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Given that k is a positive integer and we need to find what is the remainder when \((k + 2)(k^3 – k)\) is divided by 6

Let's simplify \((k + 2)(k^3 – k)\)

\((k + 2)(k^3 – k)\) = \((k + 2)(k*(k^2 – 1))\) = k*(k+2)*\((k^2 - 1^2)\) = k*(k+2)*(k-1)*(k+1) = (k-1)*k*(k+1)*(k+2)
= Product of four consecutive numbers

Theory: Product of n consecutive numbers will always be divisible by n!

=> Product of four consecutive numbers will always be divisible by 4! (=1*2*3*4 = 24)
=> Product of four consecutive numbers will always be divisible by 6, as 6 is a factor of 24

=> The remainder when \((k + 2)(k^3 – k)\) is divided by 6 = 0

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Remainders


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Hello Bunnel,

Thank you for your answer, can you please explain why the product of 4 consecutive integers will always be divisible by 4?

Thank you

Hi nourelhoudask,

The concept that you are asking about is based on a couple of Number Property rules - and you can prove it rather easily if you want to by TESTing VALUES.

For a positive integer to be 'evenly divisible by 4' means that the integer is a 'multiple of 4'; in other words, you have to be able to find a '4' when you prime-factor that number (meaning that you have to find at least two 2s) in its prime-factorization.

For example, 12 is evenly divisible by 4 because when you prime-factor 12, you get (2)(2)(3).
In that same way, 15 is NOT evenly divisible by 4 because when you prime-factor 15, you get (3)(5) ---> we don't have the two 2s that we need.

When you multiply four consecutive integers together, you will ALWAYS have a multiple of 4 in that product. For example:
(1)(2)(3)(4) = (6)(4) = 24 --> a multiple of 4
(2)(3)(4)(5) = (30)(4) = 120 --> a multiple of 4
(3)(4)(5)(6) = (90)(4) = 360 --> a multiple of 4
(4)(5)(6)(7) = (210)(4) = 840 --> a multiple of4

That pattern will continue; you will just be dealing with larger and larger multiples of 4 as you go:
(5)(6)(7)(8) = (210)(8) = (420)(4) = 1680 --> a multiple of 4
(6)(7)(8)(9)
(7)(8)(9)(10)
(8)(9)(10)(11)

(9)(10)(11)(12)
Etc.

This shows that every group of four consecutive positive intergers will always include a multiple of 4, so the product of that group will - by definition - be a multiple of 4.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com
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