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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

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.

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Source: Beat The GMAT — Data Sufficiency | Sun Aug 07, 2011 2:48 am

akhilsuhag: Master | Next Rank: 500 Posts; Posts: 351; Joined: Mon Jul 04, 2011 10:25 pmFollowed by: 4 members; Thanked: 57 times

by akhilsuhag » Sun Aug 07, 2011 3:18 am

Can you please post the OA.

Thanks

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; 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.

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

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

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akhilsuhag: Master | Next Rank: 500 Posts; Posts: 351; Joined: Mon Jul 04, 2011 10:25 pmFollowed by: 4 members; Thanked: 57 times

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]

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akhilsuhag: Master | Next Rank: 500 Posts; Posts: 351; Joined: Mon Jul 04, 2011 10:25 pmFollowed by: 4 members; Thanked: 57 times

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.

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