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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
OG If k is a positive integer
This topic has expert replies
-
AbeNeedsAnswers
- Master | Next Rank: 500 Posts
- Posts: 394
- Joined: Sun Jul 02, 2017 10:59 am
- Thanked: 1 times
- Followed by:5 members
-
GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
We can test ANY POSITIVE INTEGER for k.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
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7244
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
Let’s simplify the given expression:AbeNeedsAnswers wrote: ↑Fri Aug 18, 2017 5:36 pmIf 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
(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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
