BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

If \(n\) is an integer greater than \(50,\) then the expression \((n^2-2n)(n^2-1)\) MUST be divisible by which of the

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 2058
Joined: Sun Oct 29, 2017 4:24 am
Thanked: 1 times
Followed by:5 members

If \(n\) is an integer greater than \(50,\) then the expression \((n^2-2n)(n^2-1)\) MUST be divisible by which of the

by M7MBA » Thu Oct 29, 2020 12:33 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

If \(n\) is an integer greater than \(50,\) then the expression \((n^2-2n)(n^2-1)\) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III

Answer: C

Source: Magoosh
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

Re: If \(n\) is an integer greater than \(50,\) then the expression \((n^2-2n)(n^2-1)\) MUST be divisible by which of th

by Scott@TargetTestPrep » Thu Nov 05, 2020 3:38 pm
M7MBA wrote: ↑
Thu Oct 29, 2020 12:33 pm
 If \(n\) is an integer greater than \(50,\) then the expression \((n^2-2n)(n^2-1)\) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III

Answer: C

Solution:

Simplifying, we have:

n(n - 2)(n - 1)(n + 1) = (n - 2)(n - 1)(n)(n + 1)

We see that this is a product of 4 consecutive integers, which is always divisible by 4! = 24 (and therefore, any factors of 24). Since 4 and 6 are factors of 24, then (n^2 - 2n)(n^2 - 1) is divisible by 4 and 6. However, since 18 is not a factor of 24, (n^2 - 2n)(n^2 - 1) might not be divisible by 18.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1