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If n is a positive integer and n^2 is divisible by 72, then [#permalink ]
27 Aug 2006, 08:18

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is
6
12
24
36
48

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ghantark wrote:

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is 6 12 24 36 48

B Can't be 72 since 72 is not a perfect square. First possible value is 144=12*12

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isn't it 6 ??? option A ..

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I think it is 48
I am assuming that each number should eventually be able to reach 72
48*1.5 is 72

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oops ...i was worong ... it should be 12 ..

how ..

n^2 = 72x

n = sqrt(72x)

=> n = 6* sqrt(2*x)

since n is a positive integer ... x must be atleast 2 so that sqrt can ve evaluated ...

=> n = 12 ..

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I say (D) by brute force. 72, 144, 216, 288... all all disible by 36
Kevin, do you know how to mathmatically set this up.

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GMATT73 wrote:

I say (D) by brute force. 72, 144, 216, 288... all all disible by 36 Kevin, do you know how to mathmatically set this up.

Your right my mistake.

The key here is to factor out the prime numbers

So since n^2 will always be divisible by 72.

Factor out 72=3x3x2x2x2 36= 2x2x3x3 so n^2 will be divisible by 36

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apollo168 wrote:

GMATT73 wrote:

I say (D) by brute force. 72, 144, 216, 288... all all disible by 36 Kevin, do you know how to mathmatically set this up.

Your right my mistake.

The key here is to factor out the prime numbers

So since n^2 will always be divisible by 72.

Factor out 72=3x3x2x2x2 36= 2x2x3x3 so n^2 will be divisible by 36

Nice one. Easier to make a big number smaller than to make a small number bigger. Prime factorization 101.

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the number could range from 12 to .... what is OA?

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nitinlgb wrote:

the number could range from 12 to .... what is OA?

Nope, as Apollo just demonstrated, the number has to be divisible by 3^2. 48 won't work.

The OA has been proven.

Unlike verbal problems, we can confirm answers in the

math forum

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NO this is 12...what are you guys talking about??

N^2=72M

so lets look at 72=2^3*3^2

n^2 prime factors are 2^3 . 3^2

now if N= sqrt(n^2)=2^3/2 * 3

now we all noe there is no such number as 2^3/2...i.e 2^1.5...so we need 2^2...

the least number possible is 2^2*3=12;

B it is...this is how you solve these problems....

GMATT73 wrote:

nitinlgb wrote:

the number could range from 12 to .... what is OA?

Nope, as Apollo just demonstrated, the number has to be divisible by 3^2. 48 won't work.

The OA has been proven.

Unlike verbal problems, we can confirm answers in the

math forum

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