BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

800 score free cat.

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1077
Joined: Mon Dec 13, 2010 1:44 am
Thanked: 118 times
Followed by:33 members
GMAT Score:710

800 score free cat.

by bblast » Sun Aug 07, 2011 2:48 am
If a,m and n are postive integers, is n^2a a multiple of m^a.


1>n is a multiple of m/2
2>n is a multiple of 2m

B

the way i solved it was that 2 is sufficient to say- NO. however oa says otherwise.
Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_
Top
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 351
Joined: Mon Jul 04, 2011 10:25 pm
Thanked: 57 times
Followed by:4 members

by akhilsuhag » Sun Aug 07, 2011 3:18 am
Can you please post the OA.

Thanks
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Sun Aug 07, 2011 3:22 am
bblast wrote:If a,m and n are postive integers, is n^2a a multiple of m^a.

1>n is a multiple of m/2
2>n is a multiple of 2m
Statement 1: n = k*(m/2), where k is an integer
Hence, n^(2a) = (km/2)^(2a) = [k^(2a)]*[m^(2a)]/(4^a) = [k^(2a)/(4^a)]*(m^a)*(m^a)

Now, [k^(2a)/(4^a)]*(m^a) may or may not be an integer. Hence, n^(2a) may or may not be a multiple m^a.

Not sufficient

Statement 2: n = k*(2m), where k is an integer
Hence, n^(2a) = (2km)^(2a) = [(2k)^(2a)]*[m^(2a)] = [(2k)^(2a)]*(m^a)*(m^a)

Now, [(2k)^(2a)]*(m^a) is always an integer. Hence, n^(2a) is always a multiple m^a.

Sufficient

The correct answer is B.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Sun Aug 07, 2011 3:26 am
Hi,
Is (n^2)^a = k.m^a?, where k is an integer.
=> Is (n^2/m)^a an integer.
As you are convinced that (1) is not sufficient
From(2):
n is multiple of 2m
So, n^2 is multiple of 2m
So, n^2/m is an integer
So, (n^2/m)^a an integer -> Yes
Sufficient

Hence, B
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 351
Joined: Mon Jul 04, 2011 10:25 pm
Thanked: 57 times
Followed by:4 members

by akhilsuhag » Sun Aug 07, 2011 3:28 am
IMO B is sufficient and gives a YES. Although I am horrible at quant so dont take my word for it.

The question asks if n^2a divided by m^a is an integer. Since a is an integer If we can prove that n is divisible by m it will be sufficient.

Statement 1:

Gives us n is divisible by m/2, or 2n is divisible by m.
Clearly insufficient because m can be 2 and n 7. This wud make 2n divisible by m but not n divisible by m.
Again n and m can both be 2 and in that case n is divisible by m.

So INSUFFICIENT.

Statement 2:

Gives us n is divisible by 2m. This means every n contains at least a 2 and a m when factored among other digits but we are not concerned with the others.

This in turn gives us that n is divisible by m.

Hence SUFFICIENT.

and so[spoiler] B; i hope the OA matches.[/spoiler]
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 351
Joined: Mon Jul 04, 2011 10:25 pm
Thanked: 57 times
Followed by:4 members

by akhilsuhag » Sun Aug 07, 2011 3:29 am
I was hoping if I can get some experts to comment on my approach to the problem.
Top
Post new topic Post Reply
• Page 1 of 1