Last visit was: 18 Nov 2025, 14:16 It is currently 18 Nov 2025, 14:16
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
 [27]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,062
 [27]
HOT Competition: If n is a positive integer, then n is a multiple of 25 Aug 2020, 08:00
1
Kudos
Add Kudos
26
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

30% (02:19) correct 70% (02:13) wrong based on 435 sessions

History

Date
Time
Result
Not Attempted Yet
If n is a positive integer, then n is a multiple of how many positive integers?


(1) \(n^5*n^{(-2)}\) has exactly 5 divisors between 1 and itself, both not inclusive.

(2) \(3n^2\) has exactly 8 divisors between 1 and itself, both not inclusive.


 

This question was provided by GMATWhiz
for the Heroes of Timers Competition

 

ShowHide Answer
Official Answer
A
Most Helpful Reply
User avatar
GMATinsight
User avatar
Major Poster
Joined: 08 Jul 2010
Last visit: 18 Nov 2025
Posts: 6,835
Own Kudos:
16,349
 [26]
Given Kudos: 128
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Products:
My Reviews
Expert
Expert reply
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
Posts: 6,835
Kudos: 16,349
 [26]
HOT Competition: If n is a positive integer, then n is a multiple of Updated on: 27 Aug 2020, 11:31
Originally posted by GMATinsight on 25 Aug 2020, 08:16.
Last edited by bb on 27 Aug 2020, 11:31, edited 3 times in total.
18
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
Answer: Option A

Please check the video for the step-by-step solution.

GMATinsight's Solution



Attachment:
Screenshot 2020-08-26 at 9.16.25 PM.png
Screenshot 2020-08-26 at 9.16.25 PM.png [ 1.25 MiB | Viewed 6904 times ]
General Discussion
User avatar
NiftyNiffler
User avatar
McCombs School Moderator
Joined: 26 May 2019
Last visit: 15 Aug 2021
Posts: 325
Own Kudos:
378
 [2]
Given Kudos: 151
Location: India
Schools: INSEAD Jan '20 LSE MFin "21 McCombs '23 Tepper '23
GMAT 1: 690 Q50 V33
Products:
My Reviews
Schools: INSEAD Jan '20 LSE MFin "21 McCombs '23 Tepper '23
GMAT 1: 690 Q50 V33
Posts: 325
Kudos: 378
 [2]
HOT Competition: If n is a positive integer, then n is a multiple of 25 Aug 2020, 08:29
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
If n is a positive integer, then n is a multiple of how many positive integers?


(1) n^5∗n^(−2) has exactly 5 divisors between 1 and itself, both not inclusive.

(2) 3*n^2 has exactly 8 divisors between 1 and itself, both not inclusive.

st1) n^3 has 7 divisors (all inclusive). Thus n^3 can't be factorized to multiple prime numbers. n^3 can be represented as x^6 where x is a prime number.

Hence, n = x^2 where x is a prime number and factors of n = 3 SUFFICIENT

st2) this is a bit tricky. 3*n^2 has total 10 divisors. So, n can either be a perfect square of a prime number (3x^4) or 3*3^8. In such cases, factors of n can be either 3 or 5. So INSUFFICIENT.

The answer should be A
User avatar
hiranmay
User avatar
Joined: 12 Dec 2015
Last visit: 22 Jun 2024
Posts: 459
Own Kudos:
560
 [3]
Given Kudos: 84
Posts: 459
Kudos: 560
 [3]
HOT Competition: If n is a positive integer, then n is a multiple of 25 Aug 2020, 08:51
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
If n is a positive integer, then n is a multiple of how many positive integers?


(1) \(n^5*n^{-2}\) has exactly 5 divisors between 1 and itself, both not inclusive. --> suff: \(n^5*n^{-2}\) has 5 divisors excluding 1 & itself i.e. \(n^3\) has 5+2=7=1*7 divisors including 1 & itself, so \(n^3 = p^6\), where p is a prime number => n= p^2. so n is a multiple of 2+1=3 +ve integers

(2) \(3n^2\) has exactly 8 divisors between 1 and itself, both not inclusive. --> insuff: \(3n^2\) has 8 divisors excluding 1 & itself i.e. \(3n^2\) has 8+2=10=1*10=2*5 divisors including 1 & itself, so \(3n^2 = a^9\) or \(3n^2 = b^1*c^4\), or \(3n^2 = d^4*e^1\), where a,b,c,d,&e are prime numbers => if a=3, then \(n^2 = a^8\) or if b=3, then \(n^2 = c^4\), or if d=3, \(n^2 = d^3*e^1\)=> last one not possible, because n is a +ve integer, so \(n = a^4\) or \(n = c^2\), so when \(n = a^4\), n is a multiple of 4+1=5 +ve integers, but when \(n = c^2\), n is a multiple of 2+1=3 +ve integers. so no definite answer from (2)

Answer: A

Posted from my mobile device
User avatar
MayankSingh
User avatar
Joined: 08 Jan 2018
Last visit: 20 Dec 2024
Posts: 289
Own Kudos:
274
 [1]
Given Kudos: 249
Location: India
Concentration: Operations, General Management
Schools: McDonough '25 (A) Marshall '25 (D) Tepper '25 (WA$) Kenan-Flagler '25 (A) ISB '25 (WL) IIMC '25 (A) Owen '26 (A$$$$)
GMAT 1: 640 Q48 V27
GMAT 2: 730 Q51 V38
GPA: 3.9
WE:Project Management (Manufacturing)
Products:
My Reviews
Schools: McDonough '25 (A) Marshall '25 (D) Tepper '25 (WA$) Kenan-Flagler '25 (A) ISB '25 (WL) IIMC '25 (A) Owen '26 (A$$$$)
GMAT 2: 730 Q51 V38
Posts: 289
Kudos: 274
 [1]
HOT Competition: If n is a positive integer, then n is a multiple of 25 Aug 2020, 09:53
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
IMO A

If n is a positive integer, then n is a multiple of how many positive integers?

Note: n is a multiple of how many positive integers = No. of factor of n
If n = p^a x q^b x r^c, where p,q,r are prime numbers & a,b,c integers, then No. of factors of n= (a+1)(b+1)(c+1)

(1) n^5∗n^(−2) has exactly 5 divisors between 1 and itself, both not inclusive.

n^5∗n^(−2) = n^3 has 7 factors in total.
no. of factors = (a+1)(b+1)(c+1) = 7 = 7x1x1 => a=6 , b=c=0
So, n^3 must be of the form p^6 i.e n^3 = p^6 => n = p^2 (where p = prime number)
Since, n=p^2, so No. of factors of n=(2+1) =3 [Including 1 & no. itself]
Sufficient

(2) 3n^2 has exactly 8 divisors between 1 and itself, both not inclusive.

3n^2 has 10 factors in total.
Therefore, (a+1)(b+1)= 10= 10x1 = 2x5

Case 1: a=9 , b=0
In this case 3n^2 is of form p^9 , This is possible when n=3^4 , then 3n^2 = 3 x (3^4)^2 = 3^9
Therefore, n=3^4 has (4+1)= 5 Factors

Case 2: a=1 , b=4
In this case 3n^2 is of form p x q^4 , This is possible when p=3 & q=(Any prime no. except 3)^2
Here, n^2 = q^4 => n = q^2 (q = prime no. other than 3)
No. of factors of n = (2+1) = 3

Two possible values possible, therefore Insufficient

Posted from my mobile device
Moderators:
Bunuel
Math Expert
105355 posts
kingbucky
496 posts