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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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 x divisible by 3?

Expert replies
Source: — Data Sufficiency |

by Jay@ManhattanReview » Mon Mar 05, 2018 2:32 am
ardz24 wrote:Is x divisible by 3?

(1) x + y is divisible by 3.
(2) x - y is divisible by 3.

What's the best way to determine which statement is sufficient? Can any experts help?
This question can be dealt with an ease by choosing smart numbers for x and y.

(1) x + y is divisible by 3.

Case 1: Say x = 3 and y =0, then x + y = 3 + 0 = 3. We see that x + y and x are divisible by 3. The answer is Yes.
Case 2: Say x = 4 and y = 2, then x + y = 5 + 2 = 6. We see that though x + y is divisible by 3, x is NOT. The answer is No.

No unique answer. Not sufficient.

(2) x - y is divisible by 3.

Case 1: Say x = 3 and y =0, then x - y = 3 - 0 = 3. We see that x - y and x are divisible by 3. The answer is Yes.
Case 2: Say x = 5 and y = 2, then x - y = 5 - 2 = 3. We see that though x - y is divisible by 3, x is NOT. The answer is No.

No unique answer. Not sufficient.

(1) and (2) together

You cannot find a pair of integers such that x + y and x - y are each divisible by 3, and x is not. Thus, x is divisible by 3. The answer is Yes. Sufficient.

Let's take an algebraic route to understand this.

Say x + y = 3k, and x - y = 3q, where k and q are any integers

Thus, x = 3[(k + q)/2]; and y = 3[(k - q)/2]

We see that x is a multiple of 3. Sufficient.

The correct answer: C

Hope this helps!
_________________
Manhattan Review GMAT Prep

Locations: New York | Jakarta | Nanjing | Berlin | and many more...

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

by Brent@GMATPrepNow » Mon Mar 05, 2018 7:27 am
Jay@ManhattanReview wrote:
ardz24 wrote:Is x divisible by 3?

(1) x + y is divisible by 3.
(2) x - y is divisible by 3.

What's the best way to determine which statement is sufficient? Can any experts help?
This question can be dealt with an ease by choosing smart numbers for x and y.

(1) x + y is divisible by 3.

Case 1: Say x = 3 and y =0, then x + y = 3 + 0 = 3. We see that x + y and x are divisible by 3. The answer is Yes.
Case 2: Say x = 4 and y = 2, then x + y = 5 + 2 = 6. We see that though x + y is divisible by 3, x is NOT. The answer is No.

No unique answer. Not sufficient.

(2) x - y is divisible by 3.

Case 1: Say x = 3 and y =0, then x - y = 3 - 0 = 3. We see that x - y and x are divisible by 3. The answer is Yes.
Case 2: Say x = 5 and y = 2, then x - y = 5 - 2 = 3. We see that though x - y is divisible by 3, x is NOT. The answer is No.

No unique answer. Not sufficient.

(1) and (2) together

You cannot find a pair of integers such that x + y and x - y are each divisible by 3, and x is not. Thus, x is divisible by 3. The answer is Yes. Sufficient.

Let's take an algebraic route to understand this.

Say x + y = 3k, and x - y = 3q, where k and q are any integers

Thus, x = 3[(k + q)/2]; and y = 3[(k - q)/2]

We see that x is a multiple of 3. Sufficient.

The correct answer: C
For the answer to be C, we need some sort of restriction that states x and y are integers.
Otherwise, non-integer pairs of values such as x = 4.5 and y = 1.5 satisfy both statements.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

by Vincen » Fri Mar 09, 2018 8:05 am
ardz24 wrote:Is x divisible by 3?

(1) x + y is divisible by 3.
(2) x - y is divisible by 3.

What's the best way to determine which statement is sufficient? Can any experts help?
Hello. This is how I'd solve it:

(1) x+y is divisible by 3.

If we take x=3 and y=6 we have that x+y=9 and 9 is divisible by 3. This could implie that x must be divisible by 3; but is we take x=1 and y=5 then x+y=6 and in this case x is not divisible by 3. Therefore, this statement is INSUFFICIENT.

(2) x-y is divisible by 3.

The same as above, take first x=6 and y=3, then take x=5 and y=2. INSUFFICIENT.

Now, using both statements together $$x+y=3\cdot k$$ $$x-y=3\cdot p$$ this implies that $$2x=3k+3p\ \Leftrightarrow\ \ \ x=\frac{3\left(k+p\right)}{2}$$ and this last number is divisible by 3 always.

Hence, this case is SUFFICIENT.

The correct option is C.
Join the discussion