Bunuel
If x is an integer such that x > 0, is it true that \(\sqrt{x}\) is an integer?
(1) \(9* \sqrt{16x}\) is an integer.
(2) \(\sqrt{6x}\) is not an integer.
Kudos for a correct solution. 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).