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

Question

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

Bunuel · Accepted Answer

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).

BrentGMATPrepNow · Answer

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)( 1 ³ –  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)( 2 ³ –  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)( 3 ³ –  3 )
= (5)(24)
= 120
When we divide 120 by 6 we get 20 with remainder  0 
Answer: still A

And so on...

RELATED VIDEO
 https://www.youtube.com/watch?v=XCxgIR042tE

sashiim20 · Answer

(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).

ydmuley · Answer

(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

JeffTargetTestPrep · Answer

Let’s simplify the given expression:

(k + 2)(k^3 – k) = (k + 2)  = (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

EMPOWERgmatRichC · Answer

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:  A

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

dave13 · Answer

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

Bunuel · Answer

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

GMATinsight · Answer

Answer: Option A

Video solution by  GMATinsight

https://www.youtube.com/watch?v=cNhrAFm8Ieg

CrackverbalGMAT · Answer

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

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=P0S6j2uTaj4

nourelhoudask · Answer

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

EMPOWERgmatRichC · Answer

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

ablatt4 · Answer

Sometimes the fastest solution is to just hunt a pattern. We know K is a positive integer so lets try some.

If k=1
3(0)/6 = 0

If k=2
4(6)/6=4

If k=3
5(24)/6=20

None of these have any remainders, 0 is the answer.

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

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

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