Last visit was: 12 Jul 2024, 16:41 It is currently 12 Jul 2024, 16:41

Close

GRE Prep 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 GRE 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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [13]
Given Kudos: 25479
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11864 [12]
Given Kudos: 136
Send PM
General Discussion
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [1]
Given Kudos: 25479
Send PM
avatar
Manager
Manager
Joined: 22 Jan 2020
Posts: 120
Own Kudos [?]: 234 [2]
Given Kudos: 10
Send PM
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]
1
1
Bookmarks
since n^2 is divisible by 72
this means 72 is a factor of n^2

That is, n^2 can be written as

n^2=72*m (where m is some integer)

therefore
n=(72*m)^0.5 = (2^3 * 3^2 * m)^0.5

Since n is a positive integer there must be at least one factor of 2 in m, b/c if there wasn't then (2^3)^0.5 would not be an integer and therefore n wouldn't be an integer.

Therefore
n=(72*m)^0.5 = (2^3 * 3^2 * m)^0.5 = (2^4 * 3^2 * (m/2))^.5
= (2^4)^0.5 * (3^2)^0.5 * (m/2)^0.5
=2^2 * 3 * (m/2)^0.5
= 4*3* (m/2)^0.5
=12* (m/2)^0.5

Therefore 12 is the greatest positive integer that must divide n.

Final Answer: B
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Own Kudos [?]: 360 [3]
Given Kudos: 64
GRE 1: Q160 V152
Send PM
If n is a positive integer and n^2 is divisible by 72, then [#permalink]
2
1
Bookmarks
Divisor \(72=2^3*3^2\) of divident \(n^2\) can be at least multiplied by \(2\) to contain even powers for the perfect square. Hence, the least perfect number should be \(n^2=2^4*3^2\) and \(\sqrt{n^2}\)=\(|n|\)=\(|2|^2*|3|\). The largest positive integer is \(12\). Answer must be B

Carcass wrote:
If n is a positive integer and \(n^2\) is divisible by 72, then the largest positive integer that must divide n is

A. 6
B. 12
C. 24
D. 36
E. 48
Intern
Intern
Joined: 15 Oct 2022
Posts: 14
Own Kudos [?]: 18 [1]
Given Kudos: 70
Location: India
Concentration: Technology, Finance
GPA: 3.52
WE:Analyst (Computer Software)
Send PM
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]
1
We can use the hit-and-trial approach as well to solve it quickly under time pressure if this kind of question comes in the actual GRE test.
If we just think of a known perfect square that is greater than 72 and is divisible by 72, then it would be 144=12^2.
Then it is quite obvious that the largest number which divides n can be 12 only (This approach can be used if you have less than a minute to solve this question).
Intern
Intern
Joined: 21 Jan 2021
Posts: 23
Own Kudos [?]: 6 [0]
Given Kudos: 0
Send PM
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]
Have to make it make sense as a square first- fix the exponent. Then take the square root:

Prep Club for GRE Bot
[#permalink]
Moderators:
GRE Instructor
46 posts
GRE Forum Moderator
17 posts
Moderator
1090 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne