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.

Divisibility problem

Expert replies
by daygoballa » Sat May 23, 2009 3:47 pm
x is divisible by 144. If cube root of "x" is an integer, then which of the following is cube root of "x" definitely divisible by? (choose all that apply)

a. 4
b. 8
c. 9
d. 12

I simply just don't understand what the problem is asking. Can someone please interpret what the question is asking?

thank you!
Join the discussion
Source: — Problem Solving |

by ssmiles08 » Sat May 23, 2009 5:05 pm
I am not sure I am a little weak in these areas but my approach is that if a cube root of x is an integer, then the prime factors of x must occur in sets of 3.

Here x is divisible by 144. 144 has 2*2*2*2*3*3.

2 occurs 4 times and 3 occurs only 2 times. two more 2's are needed to make this two sets of 2's and one more 3 is needed to make it one set of 3.

so 2*2*3 = 12

So I would say x is definitely divisible by 12 and since 4 is also a divisor of 12, it would be divisible by 4 as well.

What is the OA
Join the discussion

by daygoballa » Sun May 24, 2009 1:34 am
the answer is 4 and 12...thank you!
Join the discussion

by alexdallas » Sat Aug 15, 2009 4:54 pm
tks baws
best explanation ive seen on this problem.
Join the discussion