GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Jul 2018, 14:30

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

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

If m is a positive integer and m^2 is divisible by 48, then

Author Message
TAGS:

Hide Tags

Intern
Joined: 18 Sep 2011
Posts: 8
Concentration: Strategy, General Management
GMAT Date: 11-29-2011
WE: Corporate Finance (Retail)
If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

10 Oct 2011, 12:42
2
6
00:00

Difficulty:

25% (medium)

Question Stats:

72% (01:00) correct 28% (00:59) wrong based on 488 sessions

HideShow timer Statistics

If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA
Math Expert
Joined: 02 Sep 2009
Posts: 46991
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 03:25
1
4
M3tm4n wrote:
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

Posted from my mobile device

If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?
(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

m^2 is a positive perfect square divisible by 48, the least such perfect square is 144=48*3=12^2. Thus the least value of m is 12, which means that m in any case must be divisible by 12.

Similar questions to practice:
http://gmatclub.com/forum/properties-of ... 90523.html
http://gmatclub.com/forum/if-m-and-n-ar ... 08985.html
http://gmatclub.com/forum/property-of-i ... 04272.html
http://gmatclub.com/forum/if-x-and-y-ar ... 00413.html
http://gmatclub.com/forum/number-properties-92562.html
http://gmatclub.com/forum/can-someone-a ... 92066.html
http://gmatclub.com/forum/og-quantitative-91750.html
http://gmatclub.com/forum/division-factor-88388.html
http://gmatclub.com/forum/if-5400mn-k4- ... 09284.html

Hope it helps.
_________________
General Discussion
Manager
Joined: 07 Oct 2010
Posts: 154

Show Tags

10 Oct 2011, 12:56
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.
Manager
Joined: 21 Aug 2010
Posts: 180
Location: United States
GMAT 1: 700 Q49 V35

Show Tags

10 Oct 2011, 13:05
1
arcanis2000 wrote:
Q: If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

Can someone please explain this one? Knewton's explanation was not useful to me. Thx
OA

M^2 is divisible by 48 so M^2 must be multiple of 48.
If the value of M is Multiples of 12 then it will satisfy the condition. If we If M is 12 or 24 or 36 then it ans is D but if M = 48 then answer should be 16.

Is the question right? Or am i missing some thing?
_________________

-------------------------------------

Manager
Joined: 21 Aug 2010
Posts: 180
Location: United States
GMAT 1: 700 Q49 V35

Show Tags

10 Oct 2011, 13:07
vyassaptarashi wrote:
Well m^2 is divisible by 48

if we brake 48 we will get 4*4*3

now we know that m is an integer therefore, m^2 is completely divisible by 48 then to get the square number that will be divided by 48 we need to multiply 48 by 3
so that we will get 4*4*3*3 ....
square root of the above number will be 4*3 = 12
thus m = 12 and the greatest number that will divide 12 is 12 itself ...thus the answer.

This explanation looks fine but why M shouldn't be 48. There is no mention in the question about M's maximum limit.
_________________

-------------------------------------

Manager
Joined: 07 Oct 2010
Posts: 154

Show Tags

10 Oct 2011, 13:22
1
Well, the largest integer, which will divide m, has asked in the question. This means there will be no other integer that will divide m is larger than what the answer is.
Also, It is not asked us that which of the following will be the largest integer that will divide m....

Thus m can not be 48 because the largest integer that divides 48 will be 48 ...which is not in the options, hence can not be correct. 16 is also not correct because it is the largest number in the options but larger that 16 divisors are also present.
Manager
Status: D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Joined: 12 Apr 2011
Posts: 207
Location: Kuwait
Schools: Columbia university
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

10 Oct 2011, 15:45
ok, I will try to explain this as best as possible.
the following is a tree diagram, or columns. hope it is easy to see it

m^2____m__ 2 2
____m__ 2 2 3

so, 2 * 2* 3= 12

In another words,
find out the prime factors of 48, ( 2,2,2,2,3)

, you know that M^2 is m repeated twice, so you

should distribute the the prime factors evenly

among the 2 m's, the 3 will be left over so you

should just put it out there under one of the m's.

Now, for every columns take one common number

which is 2 for the first columns, 2 for the second

column and 3 in the last column. multiplying the

prime factors ( 2*2*3) = 12...

best of luck
_________________

Sky is the limit

Director
Joined: 01 Feb 2011
Posts: 684
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

10 Oct 2011, 17:37
1
m^2 is divisible by 48

=> 48 is a factor of m^2

=> 2^4*3 is a factor of m^2

to make the above number a perfect square m^2 must have another 3 as a factor

=> (2^4)*(3^2) must be a factor of m^2

=> (2^2)*(3^1) = 12 must be a factor of m.

Director
Joined: 01 Feb 2011
Posts: 684
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

10 Oct 2011, 19:31
its mentioned indirectly by saying m has to be an integer.

if m^2 is not a perfect square then m will not be an integer.

for m to be in an integer, m^2 has to be a perfect square.

lets say m^2=48*3 = 144 (perfect square ) => m =12 (integer)

lets say m^2 = 48(not a perfect square) = > m= sqrt(48) = 2*sqrt(12) (not an integer)

arcanis2000 wrote:

@Spidy001: The perfect square is what Knewton gave as part of the answer. In what part of this question does it suggest that M^2 has to be a perfect square? What am I missing?

Thx
Intern
Joined: 06 Nov 2011
Posts: 35
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
GPA: 3
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 01:25
I dont Get it. Factors of 48 are= 2,2,2,2 and 3 and we know that 48 is a factor of m. Therefor 48 is also a factor of m^2.
So (2^4)*3 is a factor of m and m^2. But how do you come to the solution of 12? I don't see the link between (2^4)*3 and (2^2)*(3^2). Can someone explain please?

Posted from my mobile device
Manager
Joined: 31 Jan 2012
Posts: 71
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 01:49
m^2 is a divisor of 48. M =/= divisor of 48.

Find the prime factors of 48, [2,2,2,2,3]. When you root square a number you double the amount of it's divisor, so if you root them number it should be half. So divide in half all the repeating multiples. You know there must be a 3, since 3 is a prime and no 2 integer forms a multiple of 3. Also half the 2s are gone so you have a two 2s left. 3*2*2 = 12.
Manager
Joined: 10 Jan 2010
Posts: 161
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GPA: 3
WE: Consulting (Telecommunications)
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 03:36
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...
Math Expert
Joined: 02 Sep 2009
Posts: 46991
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 03:43
1
3
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.
_________________
Manager
Joined: 10 Jan 2010
Posts: 161
Location: Germany
Concentration: Strategy, General Management
Schools: IE '15 (M)
GPA: 3
WE: Consulting (Telecommunications)
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 03:46
yeah it is clear! Thanks, i wish i could explain it always like you did now.
Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 146
Location: India
WE: Information Technology (Investment Banking)
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

09 Feb 2012, 07:01
Bunnel thanks for the explanation
Manager
Joined: 07 May 2015
Posts: 93
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

21 Jul 2015, 13:36
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

Current Student
Joined: 20 Mar 2014
Posts: 2642
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

21 Jul 2015, 14:08
neeraj609 wrote:
Bunuel wrote:
MSoS wrote:
Am i right if i say:
M^2 must be a multiple of 48. 48 in prime factors: 2^4 * 3^1. Thus at least the interger must have one 3 and of course a few 2´s

Greated integer which is divisible by m and an factor of m^2 would be 12...

Algebraic way: $$m^2=48*k=2^4*3*k$$, where $$k$$ is some positive integer. $$m=\sqrt{2^4*3*k}=2^2*\sqrt{3k}$$ --> the least value of $$k$$ for which $$m$$ is an integer (hence the least value of $$m$$) is for $$k=3$$ --> $$m=2^2*\sqrt{3*3}=12$$, hence $$m$$ in any case is divisible by 12..

Hope it's clear.

Thanks alot for the help here. I have a quick doubt, the question is asking for the largest value of the largest value of m. Why are we finding the least value of m. What am I missing here.

The question is asking for the largest values from the given options. You are supposed to find which one of the 5 options gives you the correct answer. You are confusing between finding the "maximum possible value" when no options are given and "maximum possible value" out of the given options. For our use, the least value worked . If lets say the least did not work, we could have looked for a multiple of 12 (say 24 or 26 or 48 etc.).
Manager
Joined: 07 May 2015
Posts: 93
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

21 Jul 2015, 17:18
Thanks alot for the response. I was also able find the explanation from on of the another post from Bunnel.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2916
Location: United States (CA)
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

12 Dec 2017, 08:11
arcanis2000 wrote:
If m is a positive integer and m^2 is divisible by 48, then the largest positive integer that must divide m is?

(A) 3
(B) 6
(C) 8
(D) 12
(E) 16

We are given that m^2/48 = integer or (m^2)/(2^4)(3^1) = integer.

However, since m^2 is a perfect square, we need to make 48 or (2^4)(3^1) a perfect square. Since all perfect squares consist of unique prime, each raised to an even exponent, the smallest perfect square that divides into m^2 is (2^4)(3^2) = 144.

Thus, m^2/144 = integer

Since m^2 is divisible by 144, we see that the largest value that divides m is 12.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11957
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: If m is a positive integer and m^2 is divisible by 48, then [#permalink]

Show Tags

17 Jan 2018, 13:06
Hi All,

This question has a number of logic 'shortcuts' built into it, so if you were low on time you could logically guess the correct answer. Here's how you could do it... To start, you have to note that the answer choices are numbers and one of them IS the LARGEST positive integer that will divide into M.

Since M is an integer and M^2 is divisible by 48, you should be able to logically deduce that M will be divisible by both 2 (since 48 is even) and 3 (you can use the 'rule of 3' to quickly determine that fact). So we need a number that is divisible by BOTH 2 and 3. With that information alone, we can eliminate Answer choices A, C and E. The prompt asks for the LARGEST positive integer, so between the two remaining answers, and since 48 has so many 2s in it - and that the remaining answers are 6 and 12 - it's highly likely that the correct answer is 12 (which you can prove using the other approaches offered here).

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Re: If m is a positive integer and m^2 is divisible by 48, then   [#permalink] 17 Jan 2018, 13:06
Display posts from previous: Sort by

Events & Promotions

 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®.