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.

DS on OG

Expert replies
by magical cook » Thu Aug 03, 2006 12:47 am
I would appreciate it if someone could explain this why.... t

hanks in advance!!!
Jane

If X and Y are positive intergers such that X = 8Y + 12, what is the common divisor of X and Y?

1) X=12u where u is an integer
2) y=12z where z is an integer

The answer is B and only question 2 is correct.
Join the discussion
Source: — Data Sufficiency |

by gdhiman » Thu Aug 03, 2006 10:40 am
Good one.

putting in first data to the above equation gets us: 12u = 8y + 12 (u cannot get a common divisor from this).

putting in second data provided we get:
x = 8*12z + 12. So we know 12 divides X for sure, and we also know y is a multiple of 12. So 12 is the common divisor.

Hope i made sense. let me know if u need more clarification.
Join the discussion

by dblazquez » Thu Aug 03, 2006 11:03 am
Hey, i was also trying to solve this good one :)

Why on the statement I we can state that you cannot get a common divisor from 12u = 8y + 12, what about 2 and four?
Join the discussion

by gdhiman » Thu Aug 03, 2006 7:58 pm
dblazquez wrote:Hey, i was also trying to solve this good one :)

Why on the statement I we can state that you cannot get a common divisor from 12u = 8y + 12, what about 2 and four?
Coz what if y = 1. then 2 is not a common divisor or even 4.
Join the discussion

by dblazquez » Sat Aug 05, 2006 12:59 pm
sorry for the delay
hmm, 12u = 8y + 12, with y = 1 is 20 therefore 2 is common divisor, thats why i was confused... i cant see the mistake

daniel
Join the discussion

by gdhiman » Sat Aug 05, 2006 10:50 pm
dblazquez wrote:sorry for the delay
hmm, 12u = 8y + 12, with y = 1 is 20 therefore 2 is common divisor, thats why i was confused... i cant see the mistake

daniel
OK.. so i will try to detail it out.:

If X and Y are positive intergers such that X = 8Y + 12, what is the common divisor of X and Y?

1) X=12u where u is an integer
2) y=12z where z is an integer

In the first option, we have 2 equations:
X=12u & X=8Y + 12
Lets make Y=1, we get x=20.
So there is no common divisor of 1 and 20. (2 is not a common divisor).

In the second option, we have 2 equations:
Y=12z & X=96z + 12.
Lets make z=1, We get Y=12 and X=108.
(So now we have a common divisor for every value of z).
Join the discussion

by dblazquez » Sun Aug 06, 2006 6:51 am
Hey thanks a lot for the explanation, now i see it much clearer... i hate this number properties problem... defintely is B
Join the discussion