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.

Common Factor

Expert replies
by fightthegmat » Tue Sep 08, 2009 10:31 am
If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation
Join the discussion
Source: — Data Sufficiency |

Re: Common Factor

by gmat579 » Tue Sep 08, 2009 12:25 pm
fightthegmat wrote:If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation

IMO - A

simplifying the given equation we get 2k + 3m = t

The question asks us if 3 or 4 is a factor of 't'

From Stmt 1 - since k is a multiple of 3, 3 has to be a factor of 't' as 2k + 3m = t

From Stmt 2 - We know that m is mutiple of 3 but we dont know about k. So not sufficient.
Join the discussion

by tom4lax » Tue Sep 08, 2009 12:55 pm
I am not sure if I follow the logic 100%.

How do we get from the question stem that 3 or 4 needs to be a factor of 12?

I get the simplification to t = 2k + 3m

Statement I) t = 2(3k) + 3m
This to me means t = 6k + 3m

Statement II) t = 2k + 3(3m)
means t = 2k + 9m

Taking a quick glance and plugging in 1's for k & m, from statement 1 we get t = 9 and from 2 we get t = 11?? I dont think this is the correct way to look at it though, but it makes the most sense to me.

Whats the correct thought process here? (Sorry for detailed question, but these types of questions come up a lot and I want to be able to nail down the concept)
Join the discussion

Re: Common Factor

by fightthegmat » Tue Sep 08, 2009 7:00 pm
gmat579 wrote:
fightthegmat wrote:If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

Can you help with the explanation

IMO - A

simplifying the given equation we get 2k + 3m = t

The question asks us if 3 or 4 is a factor of 't'

From Stmt 1 - since k is a multiple of 3, 3 has to be a factor of 't' as 2k + 3m = t

From Stmt 2 - We know that m is mutiple of 3 but we dont know about k. So not sufficient.
Could you pls elaborate more on the logic, if statement 1 says k is a multiple of 3 and as to be a factor of k then why not statement which says m is a multiple of 3 so it can be a factor of t(accor to your logic)

The OA is A,but i am not clear with the explanation
Join the discussion

by bharathh » Tue Sep 08, 2009 9:09 pm
Take the original eqn and multiply by 12.

You get 2k + 3m = t

Now for t to share a factor with 12 we need to prove that it is the product of 2 and some number or 3 and some number (factors of 12)

Statement 1

If k is a multiple of 3, it can be expressed as k = 3n

so 2*3n + 3m = t

Factor 3 out

3(2n + m) = t

Sufficient as t can be expressed as the product of 3 and someone number. So it shares 3 as a common factor with 12.


Statement 2

We already know that 3m is divisible by 3. Stating that m is also divisible by 3 does not help us any. Regardless lets plug in m = 3n

2k + 3(3m) = t

Cannot do anything more. So we have no idea if t can be expressed as a product of 2 and a number or 3 and a number.

Since statement 1 is sufficient, A
Join the discussion

by tom4lax » Wed Sep 09, 2009 5:46 am
Brilliant, thanks so much. The step where we factor out the 3 to get to t = 3(2n+m) is what I was missing.

Very clear now.
Join the discussion