It is currently 29 Nov 2023, 00:02
Register
GMAT Club Tests
GMAT Club Tests
Decision Tracker
My Rewards
New posts
New comers' posts

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

What is the smallest positive integer n such that 6,480n^1/2

17 posts Go to First Unread Post
Print view
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
SORT BY:
Date
Kudos
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 90089
CAT Tests Forum Quiz Mode
What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Mar 2014, 06:22
6
Kudos
63
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

68% (02:17) correct 32% (02:09) wrong based on 1616 sessions

Hide Show timer Statistics

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

Experience a GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
GMAT Club Tests Check Our Products Try Free Tests
ShowHide Answer
Official Answer
Official Answer and Stats are available only to registered users. Register/Login.

_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
Most Helpful Reply
User avatar
PareshGmat
SVP
SVP
Joined: 27 Dec 2012
Status:The Best Or Nothing
Posts: 1566
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Mar 2014, 19:13
12
Kudos
4
Bookmarks
Answer = E. \(30^4\)

Looking at the options, we see that 5 & 30 are not perfect square roots; so option A & C can be discarded

\(6480 = 2^3 . 3^4 . 10\)

\(2^3\) is already a perfect cube; no further adjustment is required.

\(3^4\) requires \(3^2\) to make it perfect cube

10 requires \(10^2\) to make it perfect cube

So we require \(30^2\) to make the term a perfect cube

\(\sqrt{n} = 30^2\)

\(n = 30^4\) = Answer
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 90089
CAT Tests Forum Quiz Mode
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Mar 2014, 06:23
4
Kudos
5
Bookmarks
Expert Reply
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480:
  • \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube, \(\sqrt{n}\) must complete the powers of 2, 3 and 5 (which are 4, 4, and 1, respectively) to a multiple of 3. Thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2\).

Notice that in this case WILL be a perfect cube:
  • \(2^4*3^4*5*\sqrt{n}=2^4*3^4*5*(2^2*3^2*5^2)=2^6*3^6*5^3=(2^2*3^2*5)^3\).

The least value of \(n\), therefore, is:
  • \((\sqrt{n})^2=n=(2^2*3^2*5^2)^2=30^4\).

Answer: E.
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
General Discussion
User avatar
WoundedTiger
Director
Director
Joined: 25 Apr 2012
Posts: 533
Location: India
GPA: 3.21
WE:Business Development (Other)
Reviews Badge
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Mar 2014, 07:15
1
Kudos
Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4


Sol: Let's factorize 6480 and we get 6480= 3^4*2^4*5
Now we need to see for what minimum value of \(\sqrt{n}\)*6480= a^3 where a is an Integer

So from 6480 we already have 2^4*3^4*5*\(\sqrt{n}\) = (2^2)^3* (3^2)^3*(5)^3 why cause a is an integer which will need to be have the same factors which are in LHS

solving for \(\sqrt{n}\) = (2^6*3^6*5^3)/ 2^4*3^4*5 and we get

\(\sqrt{n}\)= 2^2*3^2*5^2
Or n = 2^4*3^4*5^4 or 30^4

Ans is E
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Signature Read More
User avatar
b2bt
Manager
Manager
Joined: 25 Sep 2012
Posts: 206
Location: India
Concentration: Strategy, Marketing
Schools: Weatherhead '18 (A$) Northeastern '18 (WL)
GMAT 1: 660 Q49 V31
GMAT 2: 680 Q48 V34
GMAT ToolKit User Reviews Badge
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Mar 2014, 20:59
3
Kudos
6480 = 8*810 = 8*9*90 = 8*27*3*5*2
8 and 27 are perfect cube
To make a perfect cube, we need \(3^2\)*\(5^2\)*\(2^2\)
hence\(\sqrt{n} = 3^2*5^2*2^2\)
therefore \(n^2\) = \(3^4\)*\(5^4\)*\(2^4\) = \((3*5*2)^4\) = \(30^4\)

Answer - E
Difficulty level - 600
Time Taken - 1:09
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 90089
CAT Tests Forum Quiz Mode
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  24 Mar 2014, 01:44
1
Kudos
Expert Reply
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.
_________________
New to the GMAT CLUB Forum?

My Signature Questions' Collection:


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
Signature Read More
User avatar
P
mvictor
Board of Directors
Joined: 17 Jul 2014
Posts: 2174
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '20 (WD)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE:General Management (Transportation)
GMAT ToolKit User Reviews Badge
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  19 Dec 2015, 10:53
for 6480* sqrt(n) to be perfect cube, we need to find first the prime factorization of 6480.
well, 6480 = 648*10.
648-> sum of the digits is divisible by 9, so let's divide by 9.
648=9*72=>9*9*8.
ok, so 6480=(2^4)*5*(3^4).
for the number 6480* sqrt(n) to be perfect square, sqrt(n) must be at least 2^2*5^2*3^2. since it is sqrt, then n must be 2^4*5^4*3^4 or (2*3*5)^4 or 30^4, which is E.
User avatar
D
stonecold
Alum
Joined: 12 Aug 2015
Posts: 2298
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  03 Dec 2016, 20:02
Great Official Question.
Here is what i did->
Let √n=p
now 6480 = 2^4*5*3^4
the minimum value of p must be 2^2*5^2*3^2 for the product to be a perfect cube.
Hence √n=30^2
so n=30^4
hence E

Great Question .
_________________
Give me a hell yeah ...!!!!!

MBA Recruiting:- EMPLOYMENT AND SALARY STATISTICS AT TOP B-SCHOOLS IN THE US! (2018)
MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!

Getting into HOLLYWOOD with an MBA!
The MOST AFFORDABLE MBA programs!
STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)
AVERAGE GRE Scores At The Top Business Schools!
Signature Read More
avatar
B
dineshril
Intern
Intern
Joined: 27 Apr 2015
Posts: 35
GMAT 1: 370 Q29 V13
Reviews Badge
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  25 Dec 2017, 10:32
Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4


Prime Factorice 6480
=> \(6480=(2^4*3^4*5)\)
=> Required \(6480*\sqrt{n}=perfect cube=(2^4*3^4*5)*\sqrt{n}\)
=> This set of prime factors \((2^4*3^4*5)\) make perfect cube when multiplied by (\(2^2*3^2*5^2\))

=> i.e \((2^4*3^4*5)*(2^2*3^2*5^2)\)= \((2^2*3^2*5)^3\)

=> \(6480*\sqrt{n}\)=\((2^4*3^4*5)*\sqrt{n}\)=\((2^4*3^4*5)*(2^2*3^2*5^2)\)

=> \(\sqrt{n}\)=\((2^2*3^2*5^2)\)

=> \(\sqrt{n}\)=\((30)^2\)

=> squaring both the sides we have

=> n=\((30^2)^2\)
=> n=\(30^4\)

Therefore 'E'

Thanks
Dinesh
User avatar
L
dave13
VP
VP
Joined: 09 Mar 2016
Posts: 1182
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  07 May 2018, 10:34
1
Kudos
Bunuel wrote:
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.


Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? :) please do advise :)
thank you !

by the way is there a rule .... for example for a number \(xxxxxxx\) to be divisible by number \(yyyyyyy\) \(y\) should have same number of prime numbers ? :? can you please refresh my memory with your sharp words like a clear sky lighting :)

pushpitkc hello may be you can explain the above :-)
User avatar
L
pushpitkc
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 2895
Location: India
GPA: 3.12
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  08 May 2018, 01:25
dave13 wrote:
Bunuel wrote:
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.


Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? :) please do advise :)
thank you !

by the way is there a rule .... for example for a number \(xxxxxxx\) to be divisible by number \(yyyyyyy\) \(y\) should have same number of prime numbers ? :? can you please refresh my memory with your sharp words like a clear sky lighting :)

pushpitkc hello may be you can explain the above :-)


Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is \(2^4*3^4*5\). In order for this number to be a perfect cube,
we need to multiply it with \(2^2*3^2*5^2 = 4*9*25 = 900\) because then we will have 2
sets of \(2^3, 3^3,\) and 1 set of \(5^3\) and you will have a perfect cube.

\(900 = 30^2\). Since, we need to multiply \(\sqrt{n}\) with 6480 to make it a perfect cube, the value of n is \((30^2)^2 = 30^4\)

Hope this helps you!
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18319
Location: United States (CA)
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  09 May 2018, 15:51
Expert Reply
Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4



First, let’s factor 6,480 into primes; we have:

6480 = 10 x 648 = 10 x 9 x 72 = 10 x 9 x 9 x 8 = 2 x 5 x 3^2 x 3^2 x 2^3 = 2^4 x 3^4 x 5^1

Recall that a perfect cube requires that the exponents of all the prime factors be multiples of 3. Note that we currently have 6480 = 2^4 x 3^4 x 5^1, which means that the square root of n must be 2^2 x 3^2 x 5^2. Therefore, we see that the square root of n = 30^2, so n must be (30^2)^2 = 30^4.


Answer: E
_________________
Signature Read More
User avatar
L
dave13
VP
VP
Joined: 09 Mar 2016
Posts: 1182
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  10 May 2018, 10:49
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.


Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? :) please do advise :)
thank you !

by the way is there a rule .... for example for a number \(xxxxxxx\) to be divisible by number \(yyyyyyy\) \(y\) should have same number of prime numbers ? :? can you please refresh my memory with your sharp words like a clear sky lighting :)

pushpitkc hello may be you can explain the above :-)


Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is \(2^4*3^4*5\). In order for this number to be a perfect cube,
we need to multiply it with \(2^2*3^2*5^2 = 4*9*25 = 900\) because then we will have 2
sets of \(2^3, 3^3,\) and 1 set of \(5^3\) and you will have a perfect cube.

\(900 = 30^2\). Since, we need to multiply \(\sqrt{n}\) with 6480 to make it a perfect cube, the value of n is \((30^2)^2 = 30^4\)

Hope this helps you!



hey pushpitkc :)

you say

"In order for a number to be a perfect cube, you need to have sets of 3 prime numbers." For example is 27 a perfect cube = 27 has thees 3s \(3^3\) am i correct ?

or what does "sets of 3 prime numbers" mean ?

How about perfect square ? for instance 25 has 2 prime numbers - does it meant the same as two sets of prime numbers ?
User avatar
L
pushpitkc
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 2895
Location: India
GPA: 3.12
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  10 May 2018, 10:54
1
Kudos
dave13 wrote:


hey pushpitkc :)

you say

"In order for a number to be a perfect cube, you need to have sets of 3 prime numbers." For example is 27 a perfect cube = 27 has thees 3s \(3^3\) am i correct ?

or what does "sets of 3 prime numbers" mean ?

How about perfect square ? for instance 25 has 2 prime numbers - does it meant the same as two sets of prime numbers ?


Hi dave13

That was what I meant.
To illustrate with an example, 216 (when prime factorized) gives us 2^3 and 3^3
So, we have two sets of 3 prime numbers, making 216 a perfect cube.

Hope this helps you!
User avatar
L
dave13
VP
VP
Joined: 09 Mar 2016
Posts: 1182
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  10 May 2018, 13:06
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
SOLUTION

What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

First make prime factorization of 6,480: \(6,480=2^4*3^4*5\).

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt{n}\) must complete the powers of 2, 3 and 5 to a multiple of 3, thus the least value of \(\sqrt{n}\) must equal to \(2^2*3^2*5^2=900\). Thus the least value of \(n\) is \(900^2=30^4\).

Answer: E.


Hi Bunuel, i cant underastand the wording of this sentence must complete the powers of 2, 3 and 5 to a multiple of 3, .... to a multiple of 3 ...how can i interpret this info ? :) please do advise :)
thank you !

by the way is there a rule .... for example for a number \(xxxxxxx\) to be divisible by number \(yyyyyyy\) \(y\) should have same number of prime numbers ? :? can you please refresh my memory with your sharp words like a clear sky lighting :)

pushpitkc hello may be you can explain the above :-)


Hey dave13

In order for a number to be a perfect cube, you need to have sets of 3 prime numbers.

6480, when prime-factorized is \(2^4*3^4*5\). In order for this number to be a perfect cube,
we need to multiply it with \(2^2*3^2*5^2 = 4*9*25 = 900\) because then we will have 2
sets of \(2^3, 3^3,\) and 1 set of \(5^3\) and you will have a perfect cube.

\(900 = 30^2\). Since, we need to multiply \(\sqrt{n}\) with 6480 to make it a perfect cube, the value of n is \((30^2)^2 = 30^4\)

Hope this helps you!


pushpitkc YAY! :) i have one question :-)

you say: "when prime-factorized is \(2^4*3^4*5\). In order for this number to be a perfect cube,
we need to multiply it with \(2^2*3^2*5^2 = 4*9*25 = 900\)"

so when i multiply \(2^4*3^4*5\) by \(2^2*3^2*5^2\) = i get this:

\(2^4*2^2 = 2^6\)

\(3^4*3^2 = 3^6\)

\(5^1*5^2 = 5^3\)

\(2^6*3^6*5^3\)


so how do you get this "then we will have 2
sets of \(2^3, 3^3,\) and 1 set of \(5^3\)" :?

i mean you have \(2^3\) , \(3^3\)and \(5^3\) whereas i get \(2^6,\) \(3^6\), \(5^3\) :?

can you please explain :)
User avatar
L
pushpitkc
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 2895
Location: India
GPA: 3.12
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  10 May 2018, 23:58
1
Kudos
dave13 wrote:

pushpitkc YAY! :) i have one question :-)

you say: "when prime-factorized is \(2^4*3^4*5\). In order for this number to be a perfect cube,
we need to multiply it with \(2^2*3^2*5^2 = 4*9*25 = 900\)"

so when i multiply \(2^4*3^4*5\) by \(2^2*3^2*5^2\) = i get this:

\(2^4*2^2 = 2^6\)

\(3^4*3^2 = 3^6\)

\(5^1*5^2 = 5^3\)

\(2^6*3^6*5^3\)


so how do you get this "then we will have 2
sets of \(2^3, 3^3,\) and 1 set of \(5^3\)" :?

i mean you have \(2^3\) , \(3^3\)and \(5^3\) whereas i get \(2^6,\) \(3^6\), \(5^3\) :?

can you please explain :)


Hi dave13

This is a basic rule of exponents \(a^{m+n} = a^m * a^n\)

Coming back to the question, where you have \(2^6\) and \(3^6\), we can convert this
\(2^6 = 2^{3+3} = 2^3*2^3\) | \(3^6 = 3^{3+3} = 3^3*3^3\)

This is why I wrote that we have two sets of 2^3 and 3^3. Hope this clears your confusion.
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 30697
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink] New post  12 Sep 2023, 03:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources
Signature Read More
GMAT Club Bot
gmatclubot
Re: What is the smallest positive integer n such that 6,480n^1/2 [#permalink]
12 Sep 2023, 03:22
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
Moderators:
Bunuel
Math Expert
90088 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3141 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions