If x is an integer such that x > 0, is it true that x is an integer?

GMATinsight, how can \(\sqrt{x}\) =1/2 when it is given that x = integer?

(1) \(9* \sqrt{16x}\) is an integer.

\(9* 4\sqrt{x}\) is an integer...
therefore \(\sqrt{x}\) will be an integer..suff

2) \(\sqrt{6x}\) is not an integer...
as \(\sqrt{6}\) is not an integer, so \(\sqrt{x}\) can be an integer or non integer and still satisfy the condition... insuff

ans A

from(1)
9∗√(16x) is an integer. => 9*4√x is an Integer=> 36√x is an integer=> this is only possible when x is a perfect square and hence √x belongs to Integer Sufficient

from(2)
√(6x) is not an integer=> it can be x is a perfect square hence √(6x) is not or it can be x is not a perfect square .There can be multiple scenarios and multiple answers. hence Not Sufficient

Answer: A

800score Official Solution:

Let’s evaluate each of the statements independently first. The question assumes that x is a positive integer. In order for (1) to be true, it should be obvious to the reader that x must be a perfect square. In other words we can break down (1), using the well-known properties of roots. Now, if 4√x is an integer, it must be true that √x is in fact an integer; there is no way that the product of an integer and an irrational number can give an integer. So, statement (1) is sufficient.

Now, let’s look at Statement (2). Sometimes it will answer the question, other times it won’t. For example, if x = 9 then √54 is not an integer. Yet, √9 = 3 is an integer. On the other hand, if x = 2 then √12 is not an integer and neither is√2. So, Statement (2) is insufficient, and the only reasonable answer choice is (A).