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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Which of the following could be true of at least some of the

Expert replies
by AAPL » Thu Sep 06, 2018 5:55 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

Magoosh

Which of the following could be true of at least some of the terms of the sequence defined by $$b_n=(2n-1)(2n+3)$$

I. divisible by 15
II. divisible by 18
III. divisible by 27

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, III

OA D.
Join the discussion
Source: — Problem Solving |

AAPL wrote:Magoosh

Which of the following could be true of at least some of the terms of the sequence defined by b_n=(2n-1)(2n+3) ?

I. divisible by 15
II. divisible by 18
III. divisible by 27

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, III
From the fact that n is the index in the sequence, we know (implicitly) that n is a positive integer.

I. May the expression (product) given have (at least) one 3 AND one 5?
Sure. Take n=3, for instance, so that b_3 = 5*9 and we have our needs satisfied!

II. May the expression (product) given have (at least) one 2 AND two 3´s?
NO. Reason: (2n-1) and (2n+3) are both ODD numbers (for any integer value of n).

III. May the expression (product) given have (at least) three 3´s?
Sure. Take n=14, because (2n-1) = 27 , therefore (2n-1)*(2n+3) = 27*(positive integer) ...

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion

by Scott@TargetTestPrep » Sat Apr 13, 2019 5:49 pm
AAPL wrote:Magoosh

Which of the following could be true of at least some of the terms of the sequence defined by $$b_n=(2n-1)(2n+3)$$

I. divisible by 15
II. divisible by 18
III. divisible by 27

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, III

OA D.
We see that no matter what n is, (2n - 1) and (2n + 3) will each be odd. So b_n is odd, and it can never be divisible by 18, which is even. Now, if one of the factors of b_n is 15 (for example, if n = 8, we have 2n - 1 = 15), then b_n will be divisible by 15. Similarly, if one of the factors of b_n is 27 (for example, if n = 12, we have 2n + 3 = 27), then b_n will be divisible by 27. Therefore, we see that b_n can be divisible by 15 and 27 for some values of n, but it can't be divisible by 18 for any values of n.

Answer: D

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
Join the discussion