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
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
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
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
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.

715+ 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.

Is the integer x divisible by 6?

Expert replies
Source: — Data Sufficiency |

by Jay@ManhattanReview » Wed Nov 27, 2019 10:55 pm
BTGmoderatorDC wrote:Is the integer x divisible by 6?

(1) x + 3 is divisible by 3
(2) x + 3 is an odd number

OA C

Source: GMAT Prep
Let's take each statement one by one.

(1) x + 3 is divisible by 3

Case 1: Say x = 6, then x + 3 = 9, divisible by 3. We see that x is divisible 6. The answer is yes.
Case 2: Say x = 9, then x + 3 = 12, divisible by 3. We see that x is not divisible 6. The answer is no.

No unique answer. Insufficient.

(2) x + 3 is an odd number.

Given that x + 3 is odd, x must be even. If x = 6, the answer is yes. However, if x = 8, the answer is no.

No unique answer. Insufficient.

(1) and (2) together

From (2), say x = 2n, where n is a positive integer. By taking x = 2n, we insured that x is even.

From (1), we know that (x + 3) = (2n + 3) is divisible by 3.

Or, 2n/3 + 3/3 = integer. Thus, 2n must be divisible by 3, or n must be divisible by 3. This implies that x = 2n is divisible by 2*3 = 6. Sufficient.

The correct answer: E

Hope this helps!

-Jay
_________________
Manhattan Review GRE Prep

Locations: GRE Classes USF | GRE Prep Course Hong Kong | GRE Prep Austin | LSAT Prep Course NYC | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Join the discussion

by swerve » Thu Nov 28, 2019 10:53 am
BTGmoderatorDC wrote:Is the integer x divisible by 6?

(1) x + 3 is divisible by 3
(2) x + 3 is an odd number

OA C

Source: GMAT Prep
1)\( x + 3\) is divisible by 3

\(x+3 = 3p\)
\(x = 3p - 3\)

\(p=-1, x=-6\)
\(p=0, x=-3\)
\(p=1, x=0\)
\(p=2, x=3\)
\(p=3, x=6\)
\(p=4, x=9\)
\(p=5, x=12\)

We can see that when \(p\) is odd, \(x\) will be even and divisible by 6.
Not sufficient because we don't know the value of \(p\) or \(x\). \(\Large{\color{red}\chi}\)

2) \(x + 3\) is an odd number
If \(x = 2\) then no, if \(x = 6\) then yes, in other words, we don't know if \(x\) is even or odd.
Not sufficient. \(\Large{\color{red}\chi}\)

Together, 2) \(x + 3\) is odd, so 1) \(x + 3 = 3p\) is odd, therefore \(3p\) is odd and \(x\) will be divisible by 6. \(\Large{\color{green}\checkmark}\)
Join the discussion