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

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If n is a positive integer and n^2 is divisible by 72, then [#permalink] New post 16 Sep 2006, 09:57
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If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A)6
B)12
C)24
D)36
E)48


I am confused what is being asked here.

Please explain your answers
Thanks
Heman

Last edited by heman on 16 Sep 2006, 10:05, edited 1 time in total.
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 [#permalink] New post 16 Sep 2006, 10:01
A) n= 6 --> 6^2 not divisible by 72
B) n=12 --> 12^2 is divisible by 72
C) n=24 --> 24^2 is divisible by 72
D) n=36 --> 36^2 is divisible by 72
E) n=48 --> 48^2 is divisible by 72

Largest possible integer that must divide n should be 48

E?
Heman
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 [#permalink] New post 16 Sep 2006, 10:58
?12

n^2/72 = integer

72 = 2 * 2^2 * 3^2

sq rt (n^2/72) = n/ 2^(1/2) * 2 * 3

So n^2 is divisible by multiples of 6

First perfect square equal to or greater than 72 = 144 so n = 12

Greatest divisor of n = 12

I think this also ties in with 12 being the next whole number that is a multiple of 6 - from 2^(1/2) * 2 * 3

I'm sure one of the math fluent guys will explain further
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 [#permalink] New post 16 Sep 2006, 11:50
If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A)6
B)12
C)24
D)36
E)48

from stem ( THE LARGEST POSSIBLE POSITIVE DEVISOR OF ANY POSITIVE INTIGER IS THE INTIGER ITSELF "N")

n^2 = 72x


IF N = 48 THUS N^2 = 2304 TEST IT BY DEVIDING BY 72 = 32

E is my answer
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 [#permalink] New post 16 Sep 2006, 12:19
If we aren't given any further parameters for n other than n^2 is divisible by 72 then n could equal 12 and so is not divisible by 48. Otherwise there would be no upper limit for largest possible divisor other than the answer choices provided? The stem says the largest divisor that MUST divide n.
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 [#permalink] New post 16 Sep 2006, 12:32
MBAlad u ve got a very valid point

the ward must devide may refer to the least possible value of n^2 which is 144 and thus n = 12

if this is the case then the answer is 12 (b)

because if n is intiger and thus n^2 is a perfect square = 72x

and least possible value for x = 2 ie n^2 can never be less but can be more

so as we are given the worst case scenario ( can never be less than 144)

12 must devide but the other choices ( may be or may be not depending on the values of x in n^2 = 72x

I believe you are right and i did this silly mistake

thanks man :lol:

Last edited by yezz on 17 Sep 2006, 01:15, edited 1 time in total.
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 [#permalink] New post 16 Sep 2006, 12:46
Thanks Yezz - GMAT in T-2 days so I need some encouragement :?
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 [#permalink] New post 16 Sep 2006, 12:49
Man this should be very encouraging to you that u understand this

Paranoyd phsicotic sadish nature of the GMAT test makers :lol:

Hope you all the best and a 700+

i will wait for my 700+ at your end

you made me edit my post ..so i will not accept less than a 700+

go man go...you rock and roll too

good luck on the G DAY
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 [#permalink] New post 16 Sep 2006, 16:33
OA is B.

I was confused what was being asked. Looks like we need to find largest n that divides all n. In my approach if n=48 then it would not divide n=12

This is an OG question so might have to suck up to the poor wording of the Q

Heman
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Re: OG- Qreview#169 - PS [#permalink] New post 17 Sep 2006, 04:33
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heman wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest possible positive integer that must divide n is
A)6
B)12
C)24
D)36
E)48


I am confused what is being asked here.

Please explain your answers
Thanks
Heman


Folks, let me give a generalised answers to questions like these. Once we know the basic way to answer questions like these as yezz said we can get that 700+

Concentrate for the next 5 minutes pleaseeeeeeee

given that n^2 is divisible by 72

So n^2 = 72xk (which means 72xk must be perfect squares)
n^2 = 9x4x2xk

Clearly 9 and 4 are perfect squares. So k must be minimum 2 to make the entire number a perfect square.
SO the general form of the k will be 2 x any perfect square.
ie k = 2x p^2

So n^2 = 9x4x2x2xp^2
hence n = 3x2x2xp

So the general value of n will 12xp, which must be divisible by12

I think i am lucid,

Regards,
_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 25 Sep 2008, 00:00, edited 1 time in total.
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OG- Qreview#169 - PS [#permalink] New post 17 Sep 2006, 16:45
Thanks Cicerone,

that was a very good explaination :-D
OG- Qreview#169 - PS   [#permalink] 17 Sep 2006, 16:45
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