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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

by mitzwillrockgmat » Sat Jun 26, 2010 5:29 am
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

a. 6
b. 12
c. 24
d. 36
e. 48

Hello! Can someone help me out with this question? It's been posted a few times but none of the answers really make sense to me. Can someone give a simple, straight forward answer? Much appreciated!!!!!!!!!!

This is all I get so far:

72 = 2^3 * 3^2

so n^2/2^3 * 3^2

HELP!!!!!!!!!

answer is B
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by selango » Sat Jun 26, 2010 6:02 am
n^2 is divisible by 72

72=2*2*3*3*2

To make it the perfect square,put 2 above

72=2*2*2*2*3*3

Take common factor from each square,

2*2*3=12

Hence B
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by mitzwillrockgmat » Sat Jun 26, 2010 6:09 am
selango wrote:n^2 is divisible by 72

72=2*2*3*3*2

To make it the perfect square,put 2 above

72=2*2*2*2*3*3

Take common factor from each square,

2*2*3=12

Hence B
Thanks but a bit confused here is the below what you are saying to do...?

72 ^2 = [2^3 * 3^2 ]^2

becomes 72^2 = 2^6 * 3^4

pls elaborate, thaaaaaanks :)
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by selango » Sat Jun 26, 2010 6:22 am
mitzwillrockgmat,

Method1
---------------------

Split 72 according to prime factorization,

72=2*2*2*3*3

Since n is a perfect square,make the factors as the perfect square.

How can we make them as perfect square?

By multiplying 2 with the factors.[remember n is multiple of 72]

72=2*2*2*2*3*3

So n^2 is divisible 2*2*2*2*3*3

Take each common factor.2*2*3

Hence 12.

Method 2
-----------------------------------------------------


n^2 is divisible by 72

n^2=72q+r[q is quotient r=0]

n^2=72q

n^2=2^3*3^2*q

n^2=2^4*3^2*q[multiply by 2]

n^2=12*12*q

So 12 is the highest positive integer that divides n^2

Hope this clarifies.
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by mitzwillrockgmat » Sat Jun 26, 2010 6:31 am
selango wrote:mitzwillrockgmat,

Method1
---------------------

Split 72 according to prime factorization,

72=2*2*2*3*3

Since n is a perfect square,make the factors as the perfect square.

How can we make them as perfect square?

By multiplying 2 with the factors.[remember n is multiple of 72]

72=2*2*2*2*3*3

So n^2 is divisible 2*2*2*2*3*3

Take each common factor.2*2*3

Hence 12.

Method 2
-----------------------------------------------------


n^2 is divisible by 72

n^2=72q+r[q is quotient r=0]

n^2=72q

n^2=2^3*3^2*q

n^2=2^4*3^2*q[multiply by 2]

n^2=12*12*q

So 12 is the highest positive integer that divides n^2

Hope this clarifies.

Thanks!!! That helps but just to confirm you are saying...

since n^2 is a perfect square then its divisor must be a perfect square too.

so i need to find the factors of 72 & to square all of its factors , so we multiply the entire set of no.s by 2.

correct?
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by selango » Sat Jun 26, 2010 7:32 am
tats correct..
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by ozlemmetje » Mon May 05, 2014 12:16 pm
Hello,

there is something I don't understand:

I agree that 12 is an integer that must divide n is 12 as 72= (2^4)(3^2)is a perfect square. However (2^6)(3^2) is also a perfect square so n could also be 24 and so it is divisible by 24.

Thanks
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