Events & Promotions
| Last visit was: 06 Sep 2026, 09:46 |
It is currently 06 Sep 2026, 09:46 |
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 0712:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
05 Sep 2022, 04:37
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
56% (01:31) correct
44%
(01:50)
wrong
based on 140
sessions
History
Date
Time
Result
Not Attempted Yet
Is the positive square root of x an integer?
(1) \(x = n^3\) and n is a perfect square.
(2) \(x\) has exactly seven factors.
(1) \(x = n^3\) and n is a perfect square.
(2) \(x\) has exactly seven factors.
Part of the project: GMAT DS & PS DAILY Challenge - [TOPIC WISE] (2022) - Gain 25 Kudos & Get FREE Access to GMATClub TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GMAT - Math Book
Ultimate GMAT Quantitative Megathread
ALL YOU NEED FOR QUANT ! ! !
ALL GMAT PREP PS + DS QUESTIONS (LINKS TO ALL 825 QUESTIONS)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GMAT - Math Book
Ultimate GMAT Quantitative Megathread
ALL YOU NEED FOR QUANT ! ! !
ALL GMAT PREP PS + DS QUESTIONS (LINKS TO ALL 825 QUESTIONS)
05 Sep 2022, 06:45
Kudos
Bookmarks
(1) x=n^3 and n is a perfect square.
As n is a perfect square, let \(n = m^2\)
Which means that \(x = (m^2)^3\)
\(x = m^6\)
taking the square root of both sides will give: \(\sqrt{x} = m^3\)
As we know that \(m\) is an integer, \(m^3\) will definitely be an integer and therefore the positive square root of \(x\) will also be an integer.
Sufficient
(2) x has exactly seven factors.
As 7 is a prime number, the only possible way a number will have 7 factors is if the number has a single prime factor with a power of 6. This means that the square root of that number will be its prime factor to the power of 3, which will definitely be an integer.
Sufficient
Answer D
As n is a perfect square, let \(n = m^2\)
Which means that \(x = (m^2)^3\)
\(x = m^6\)
taking the square root of both sides will give: \(\sqrt{x} = m^3\)
As we know that \(m\) is an integer, \(m^3\) will definitely be an integer and therefore the positive square root of \(x\) will also be an integer.
Sufficient
(2) x has exactly seven factors.
As 7 is a prime number, the only possible way a number will have 7 factors is if the number has a single prime factor with a power of 6. This means that the square root of that number will be its prime factor to the power of 3, which will definitely be an integer.
Sufficient
Answer D
05 Sep 2022, 07:00
Kudos
Bookmarks
Is the positive square root of x an integer?
Two properties of Perfect Square which will be handy in solving the question
- Perfect Square is true for integers only
- The number of distinct factors of a Perfect Square is Always Odd
(1) \(x=n^3\) and n is a perfect square.
Since n is a perfect square, say n = \(a^2\), where a is an integer
then, x = \(n^3\) = \((a^2)^3\) = \(a^6\)
and \(sqrt x = sqrt(a^6) = a^3 \)-> which is an integer. Hence st1 is Sufficient.
(2) x has exactly seven factors.
since x has odd number of distinct factors (7) therefore, x is a Perfect Square
\(x = b^2\), where b is an integer
and \(sqrt x = sqrt(b^2) = b\) -> which is an integer. Hence st2 is Sufficient.
Answer: D
Two properties of Perfect Square which will be handy in solving the question
- Perfect Square is true for integers only
- The number of distinct factors of a Perfect Square is Always Odd
(1) \(x=n^3\) and n is a perfect square.
Since n is a perfect square, say n = \(a^2\), where a is an integer
then, x = \(n^3\) = \((a^2)^3\) = \(a^6\)
and \(sqrt x = sqrt(a^6) = a^3 \)-> which is an integer. Hence st1 is Sufficient.
(2) x has exactly seven factors.
since x has odd number of distinct factors (7) therefore, x is a Perfect Square
\(x = b^2\), where b is an integer
and \(sqrt x = sqrt(b^2) = b\) -> which is an integer. Hence st2 is Sufficient.
Answer: D
05 Sep 2022, 09:44
Kudos
Bookmarks
target
Is the positive square root of x an integer
√x is an integer?
#1
\(x = n^3\) and n is a perfect square
n=4,9,16,25
so x= (2^2)^3 ; ( 3^2)^3 .. 2^6 , 3^6
√x is not an integer
sufficient
#2
\(x\) has exactly seven factors.
x is prime and (x^6)
√x is not an integer
sufficient
option D
Is the positive square root of x an integer
√x is an integer?
#1
\(x = n^3\) and n is a perfect square
n=4,9,16,25
so x= (2^2)^3 ; ( 3^2)^3 .. 2^6 , 3^6
√x is not an integer
sufficient
#2
\(x\) has exactly seven factors.
x is prime and (x^6)
√x is not an integer
sufficient
option D
carcass
