Questions dealing with number properties can often be solved with very little calculations, using logic only.
We'll look for such a solution, a logical approach.

(1) Every positive integer is the square root of another integer (i.e x is the sqrt of x^2). But this doesn't mean that it also has to be the square of an integer!
Insufficient

(2) So \(\sqrt{9n}=3\sqrt{n}\) is the square of an integer meaning that \(\sqrt{n}\) can be written as k/3 for some integer k. That means that n = k^2/9 for some integer k. This is an integer only if k is divisible by 3, which it does not have to be.
Insufficient.

Combined:
So from (1) we know that n^2 must be an integer, and from (2) we know that n = k^2/9 for some integer k.
Combining, if n=k^2/9 then n^2 = k^4/81. Since this must be an integer then k^4 is divisible by 81 so k^2 is divisible by 9 and k is divisible by 3. Therefore n must be an integer.

(C) is our answer.
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You're right! My bad, fixed.
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Hi Bunuel,

Could you please provide the detailed solution for this question?

is \(n=integer^2…\sqrt{n}=integer?\)

(1) n is the square root of an integer: insufic.
\(n=\sqrt{z}\)
if \(z=16…n=\sqrt{16}=4…\sqrt{4}=integer\)
if \(z=4…n=\sqrt{4}=2…\sqrt{2}=not.integer\)

(2) \(\sqrt{9n}\) is square of an integer: insufic.
\(\sqrt{9n}=x^2…x^4=9n…n=\frac{x^4}{3^2}\)
if \(x=3…n=\frac{3^4}{3^2}=3^2=9…\sqrt{9}=integer\)
if \(x=2…n=\frac{2^4}{3^2}=16/9…\sqrt{16/9}=4/3=not.integer\)

(1&2) \(n=\sqrt{z}\) and \(n=\frac{x^4}{3^2}\); so, \(z=\frac{x^8}{3^4}=integer\) and \(x\) is a multiple of \(3\);
therefore, \(n=\frac{(3m)^4}{3^2}=3^2m^4…\sqrt{n}=3m^2=integer\); sufficient.