Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
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).
_________________
Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
[#permalink]
Show Tags
24 Jun 2017, 16:15
7
\((k + 2) (k^3 – k)\)
\((k + 2) (k (K^2 - 1))\)
\((k + 2) k (k + 1) (k - 1)\)
\((k - 1) k (k + 1) (k + 2)\)
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 _________________
"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."
Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475
If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
[#permalink]
Show Tags
Updated on: 13 Nov 2018, 09:21
3
Top Contributor
4
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
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.
Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
[#permalink]
Show Tags
09 Jan 2018, 18:28
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.
_________________
Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k
[#permalink]
Show Tags
03 Mar 2019, 21:00
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
84% (01:16) correct 
