Is x an even integer?

My thought for answering the question was:

If X^2 is divisible by 4, we can think:
X^2 = 4a, where a is a number from 1 to inf, so
sqrt(x^2) = x = 2 * sqrt(a).

Since a can be any number, sqrt(a) is not constrained to an integer. The same logic can be applied to the second statement; therefore, the answer is (E)
Cheers

Thanks for the clarification. Whenever I see a question categorized as 700+ l am much more skeptical, sometimes forgetting how the verbiage is used.

Either way, that means x^2/4 and x^4/8 must be integers. This allows for integer solutions for X such as 2,4,8, etc....however it also allows for non-integer solutions for X such as the square root of 8. (sqrt8)^2/4 = integer. (sqrt8)^4/4 = integer.

Therefore answer is E.

1) At least two possible variants: \(x = 2\) or \(x = \sqrt{12}\)
Insufficient

2) At least two possible variants: \(x = 2\) or \(x = ^4\sqrt{32}\)
Insufficient

1+2) still Insufficient: \(x = 2\) or \(x = ^4\sqrt{32}\)
Answer is E

your logic is incorrect.

In the form of x^2 /4 = a ; a must be an integer, because x^2 is divisible by 4. This results in an integer solution. Therefore a is always an integer.

X, however is not always constrained to an integer. sqrt(8)^2 / 4 = 2. Furthermore sqrt(8)^4/16 = 4.

The logic on quant problems is critical. If you can understand the basis for most of these questions, and their constraints, they're much easier to solve.

MANHATTAN GMAT OFFICIAL SOLUTION:

The problem is asking whether x is even – that is, whether x is divisible by 2.

Statement (1): NOT SUFFICIENT. It’s easy to look for “confirming” evidence by assuming the question. That’s a bad strategy. Rather, take the statement as true, and see whether you answer the question the same way every time. In particular, with a Yes/No question, be a skeptic. Try to find both a Yes case and a No case.

The statement says that x^2 is divisible by 4. That is, x^2 is a multiple of 4. List a few multiples of 4 and then answer the question using those cases.

First case: x^2 = 4 itself. So then x = 2 or –2. In both scenarios, x is indeed even, so the answer is Yes.

Second case: x^2 = 8. So then x = the positive or negative square root of 8 (√8 or –√8). Since 8 is not a perfect square, x is not an integer (the positive root is between 2 and 3). Only integers can be even, so the answer to the question is No.

Since we have a Yes case and a No case, we can stop. This statement is insufficient.

Statement (2): NOT SUFFICIENT. As in the previous analysis, you should take the statement as true, then apply different cases to the question and see whether you get the same answer each time.

The statement says that x^4 is divisible by 16. That is, x^4 is a multiple of 16. List a few multiples of 16 and then answer the question using those cases.

First case: x^4 = 16 itself. So then x = 2. This is an even integer, so the answer is Yes.

Second case: x^4 = 32. So then x = the fourth root of 32. Since 32 is not a perfect fourth power, x is not an integer, let alone an even integer. So the answer is No.

Statements (1) and (2) together: NOT SUFFICIENT. There are still Yes and No cases. In fact, the first statement implies the second statement.

If x^2 is divisible by 4, then we can write x^2 = 4n, where n is an integer.

Squaring both sides gives us x^4 = 16n^2, and since n^2 is automatically an integer if n is an integer, the second statement is also satisfied.

So the second statement adds nothing to the first statement, which is more restrictive. Every case that satisfies the first statement satisfies the second statement as well. Together, the statements are NOT sufficient.

The correct answer is E.

Hi Bunuel,
Sorry for jumping into the old thread, could you please explain more and give concrete example about the case for not choosing C?
Thank you

x = 2 satisfies both statements and gives an YES answer to the question.

\(x = \sqrt{8}\) satisfies both statements and gives a NO answer to the question (\(\sqrt{8}\) is NOT an integer and hence cannot be even, recall that only integers can be even or odd).

Hope it helps.

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