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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]
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since n^2 is divisible by 72
this means 72 is a factor of n^2

That is, n^2 can be written as

n^2=72*m (where m is some integer)

therefore
n=(72*m)^0.5 = (2^3 * 3^2 * m)^0.5

Since n is a positive integer there must be at least one factor of 2 in m, b/c if there wasn't then (2^3)^0.5 would not be an integer and therefore n wouldn't be an integer.

Therefore
n=(72*m)^0.5 = (2^3 * 3^2 * m)^0.5 = (2^4 * 3^2 * (m/2))^.5
= (2^4)^0.5 * (3^2)^0.5 * (m/2)^0.5
=2^2 * 3 * (m/2)^0.5
= 4*3* (m/2)^0.5
=12* (m/2)^0.5

Therefore 12 is the greatest positive integer that must divide n.

Final Answer: B
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If n is a positive integer and n^2 is divisible by 72, then [#permalink]
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Divisor \(72=2^3*3^2\) of divident \(n^2\) can be at least multiplied by \(2\) to contain even powers for the perfect square. Hence, the least perfect number should be \(n^2=2^4*3^2\) and \(\sqrt{n^2}\)=\(|n|\)=\(|2|^2*|3|\). The largest positive integer is \(12\). Answer must be B

Carcass wrote:
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
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Re: If n is a positive integer and n^2 is divisible by 72, then [#permalink]
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We can use the hit-and-trial approach as well to solve it quickly under time pressure if this kind of question comes in the actual GRE test.
If we just think of a known perfect square that is greater than 72 and is divisible by 72, then it would be 144=12^2.
Then it is quite obvious that the largest number which divides n can be 12 only (This approach can be used if you have less than a minute to solve this question).
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