It is currently 19 Mar 2018, 21:22

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is a positive integer and n^2 is divisible by 72, then

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44322
If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

17 Mar 2014, 00:16
4
This post received
KUDOS
Expert's post
35
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

59% (01:07) correct 41% (01:18) wrong based on 920 sessions

### HideShow timer Statistics

The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Problem Solving
Question: 169
Category: Arithmetic Properties of numbers
Page: 84
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44322
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

17 Mar 2014, 00:16
4
This post received
KUDOS
Expert's post
21
This post was
BOOKMARKED
SOLUTION

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

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

Answer: B.

Similar questions to practice:

if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html
_________________
Current Student
Joined: 25 Sep 2012
Posts: 281
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31
GMAT 2: 680 Q48 V34
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

17 Mar 2014, 04:51
3
This post received
KUDOS
1
This post was
BOOKMARKED
$$\frac{n*n}{72}$$ = $$\frac{n*n}{2*2*2*3*3}$$
This means n should be at least 2*2*3. Otherwise $$n^2$$ won't be divisible by 72

What is the largest integer that can divide n?
Let's see, largest integer that can divide 2? 2
Largest integer that can divide 3? 3
Largest integer that can divide n is n itself

Answer B
Time Taken 2:31
Difficulty level 650
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1838
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

18 Mar 2014, 02:10
1
This post received
KUDOS
Answer = B = 12

We require to find the least value of n

Say n=6, however 36 is not divisible by 72; Ignore A

Say n=12, 144 is divisible by 72

So Answer = 12
_________________

Kindly press "+1 Kudos" to appreciate

Manager
Joined: 04 Oct 2013
Posts: 161
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

18 Mar 2014, 08:00
1
This post received
KUDOS
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

Given that, n is a positive integer and n^2 is divisible by 72 means that n^2 is a perfect square that is lowest multiple of 72.

Or, the lowest multiple of 72 that is a perfect square is 144.

So, n^2 = 144q, q is some positive integer => 12 is a factor of n.

Answer: (B)

Last edited by arunspanda on 23 Mar 2014, 20:01, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 44322
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

23 Mar 2014, 07:57
Expert's post
2
This post was
BOOKMARKED
SOLUTION

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

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

Answer: B.

Similar questions to practice:

if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html
_________________
Intern
Joined: 07 Jun 2014
Posts: 17
Location: India
GMAT 1: 720 Q49 V38
GPA: 2.91
WE: Consulting (Energy and Utilities)
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

28 Jul 2014, 10:02
Bunuel wrote:
SOLUTION

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

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

Answer: B.

Similar questions to practice:

if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html

Hi,

How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??

In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.

Please help!!

Regards,
Sagar
_________________

I would rather "crash and burn" than "sulk and cry"!!

Math Expert
Joined: 02 Sep 2009
Posts: 44322
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

28 Jul 2014, 10:12
1
This post received
KUDOS
Expert's post
itsworththepain wrote:
Bunuel wrote:
SOLUTION

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

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

Answer: B.

Similar questions to practice:

if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-129929.html

Hi,

How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??

In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.

Please help!!

Regards,
Sagar

We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.
_________________
Manager
Joined: 26 Feb 2015
Posts: 123
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

14 Mar 2015, 08:56
Bunuel wrote:
itsworththepain wrote:
Bunuel wrote:
SOLUTION

We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.

I don't actually understand this either, why is it implied that we need the least value for n? I mean, as the previous poster said. 48^2 too, is divisible by 72 and 48/48 = 1.

Why is it (I mean, how do I interpret from the question that we need the least value for n)
Director
Joined: 10 Mar 2013
Posts: 581
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

21 Jun 2015, 05:16
Let's make it look easy

72= 2*2*2*3*3 we need one more 2 in order to make a perfect square we need an even number of primes --> we need one (2)
--> 72*2 =144, Take a square root of it an you'll get 12 (B)
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Last edited by BrainLab on 03 Nov 2015, 01:01, edited 1 time in total.
SVP
Joined: 08 Jul 2010
Posts: 2017
Location: India
GMAT: INSIGHT
WE: Education (Education)
If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

21 Jun 2015, 07:35
1
This post received
KUDOS
Expert's post
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

CONCEPT: Perfect Squares always have even power of Prime no. in their Prime factorised form.

$$72 = 8*9 = 2^3*3^2$$

Since power of prime numbers must be prime so $$72$$must be multiplied by minimum $$2^1$$ to make the result a perfect Square

In that Case $$n^2 = 2^4 * 3^2$$

i.e. $$n = 2^2 * 3 = 12$$

Answer: Option
[Reveal] Spoiler:
B

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Manager
Joined: 05 Jul 2015
Posts: 106
Concentration: Real Estate, International Business
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

22 Nov 2015, 20:34
I did it with exponents as other people have done.
72 = 9*8 = 3^2*2^3

n^2=72x
n = 3*2^2 (x) and X could be anything positive but 12 MUST divide n
Intern
Joined: 04 Oct 2015
Posts: 1
GMAT 1: 710 Q48 V39
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

23 Nov 2015, 03:24
I would like to emphasize one point here (which I think is not very explicit in previous answers). The following step is straightforward,
n^2 = 72 M = 2*6*6*M, where M is an integer.
For perfect square, even prime factors are needed. That is, n^2= 2*6*6*2*m (for some integer m different from M)
Since, n>0, n= (2*6)*k , where k is some other integer. That implies, n=12k where k is an integer. Meaning, n is not just a single number but any multiple of 12. This is where the question phrase, "then the largest positive integer that must divide n is.." plays the part. Since k= 1 or 2 or 3 or any other integer, no matter how large the number, "12" would be the number that would always divide "n". 6 would do the same (so would 3 or 4 or 2 hypothetically). But the largest one to always divide "n" would be "12" [among 1, 2, 3, 4, 6 or 12].
Manager
Joined: 18 Oct 2016
Posts: 139
Location: India
WE: Engineering (Energy and Utilities)
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

18 Apr 2017, 07:43
Option B

$$n^2 = 72x = 2^3*3^2*x = 2^4*3^2*w$$ : A perfect square contains Even powers of all the prime factors. [x, w & k are integers]
$$n = 2^2*3*k = 12k$$
$$k = \frac{n}{12}$$ = integer.

12 is the largest integer which must divide n evenly.
_________________

Press Kudos if you liked the post!

Rules for posting - PLEASE READ BEFORE YOU POST

Intern
Joined: 24 Jul 2017
Posts: 49
Location: India
WE: Information Technology (Computer Software)
Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]

### Show Tags

22 Sep 2017, 11:49
72 = 2^3 * 3^2

Given that n^2 is divisible by 72. So in order to satisfy this condition, n has to have a minimum value of 2^2 * 3^2 squaring which would be divisible by 72.
Hence n = 12 and largest possible value that can divide n is 12.

Kudos if it helps
Re: If n is a positive integer and n^2 is divisible by 72, then   [#permalink] 22 Sep 2017, 11:49
Display posts from previous: Sort by

# If n is a positive integer and n^2 is divisible by 72, then

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.