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
Answer: A
Source: Official guide
If k is a positive integer, what is the remainder when (k + 2)(k^3 – k) is divided by 6 ?
This topic has expert replies
-
- Legendary Member
- Posts: 1223
- Joined: Sat Feb 15, 2020 2:23 pm
- Followed by:1 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
\begin{align*}BTGModeratorVI wrote: ↑Sat Mar 28, 2020 9:57 amIf 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
Answer: A
Source: Official guide
(k+2)(k^3 – k) &=(k+2)(k(k^2−1)) \\
&= (k+2)k(k+1)(k−1).
\end{align*}
Or could be written as
\((k−1)k(k+1)(k+2)\)
Its product of 4 consecutive numbers. The product of 3 consecutive numbers is always divisible by 6, Hence given number is divisible by 6.
Therefore, the remainder will be 0, that is, A.
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Let's plug in a positive integer for k and see what happens.BTGModeratorVI wrote: ↑Sat Mar 28, 2020 9:57 amIf 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
Answer: A
Source: Official guide
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...
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7245
- 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:BTGModeratorVI wrote: ↑Sat Mar 28, 2020 9:57 amIf 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
Answer: A
Source: Official guide
(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