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If k is a positive integer, what is the remainder when (k + 2)(k^3 – k) is divided by 6 ?

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Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k) is divided by 6 ?

by Brent@GMATPrepNow » Mon Jun 08, 2020 12:39 pm
BTGModeratorVI wrote: ↑
Mon Jun 08, 2020 12:03 pm
 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
One approach is to 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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Re: If k is a positive integer, what is the remainder when (k + 2)(k^3 – k) is divided by 6 ?

by Scott@TargetTestPrep » Wed Jun 10, 2020 12:53 pm
BTGModeratorVI wrote: ↑
Mon Jun 08, 2020 12:03 pm
 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
Solution:

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 also by 6. Thus, the remainder when (k + 2)(k^3 – k) is divided by 6 is 0.

Answer: A

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