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Is x an even integer?

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Bunuel
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Is x an even integer? [#permalink] New post   15 Jun 2022, 08:30
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Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.



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BrentGMATPrepNow
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Is x an even integer? [#permalink] New post   Updated on: 15 Jun 2022, 15:59
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Bunuel wrote:
Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.


Target question: Is x an even integer?

Statement 1: \(\sqrt{x}\) is a multiple of 2
In other words, \(\sqrt{x} = 2k\) for some integer \(k\)
Now square both sides to get: \(x = 4k^2\).
This means \(x\) is a multiple of \(4\), which means \(x\) is definitely an even integer
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: \(x^2\) is a multiple of 4
There are several values of x that satisfy statement 2. Here are two:
Case a: x = √12. Here, x² = (√12)² = 12, and 12 is a multiple of 4. In this case, the answer to the target question is NO, x is not an even integer
Case b: x = 2. Here, x² = 2² = 4, and 4 is a multiple of 4. In this case, the answer to the target question is YES, x is an even integer
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Originally posted by BrentGMATPrepNow on 15 Jun 2022, 10:57.
Last edited by BrentGMATPrepNow on 15 Jun 2022, 15:59, edited 1 time in total.
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Re: Is x an even integer? [#permalink] New post   15 Jun 2022, 14:23
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BrentGMATPrepNow wrote:
Bunuel wrote:
Is \(x\) an even integer?

(1) \(\sqrt{x}\) is a multiple of 2.

(2) \(x^2\) is a multiple of 4.


Target question: Is x an even integer?

When I scan the two statements, they both feel insufficient, AND I’m pretty sure I can identify some cases with conflicting answers to the target question. So, I’m going to head straight to……

Statements 1 and 2 combined
There are several values of x that satisfy BOTH statements. Here are two:
Case a: x = √12. Here, x² = (√12)² = 12, and 12 is a multiple of 2 and a multiple of 4. In this case, the answer to the target question is NO, x is not an even integer
Case b: x = 2. Here, x² = 2² = 4, and 4 is a multiple of 2 and a multiple of 4. In this case, the answer to the target question is YES, x is an even integer
Since we can’t answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E


Hey, I reckon, based on your explanation, you have mistaken the first statement as x(squared) is a multiple of 2, instead of sqrt(x) is a multiple of 2

My answer is A

Sqrt (x) is a multiple of 2 so if sqrt(x) is anything 0,2,4,6,8,10, its square will also be divisible by 2 hence an even integer

Statement 2 is insufficient
X(squared) can be 4 (divisible by 4) which means x is 2 and YES EVEN
But if x squared is 12 (divisible by 4), x is sqrt(12), not an even number

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Re: Is x an even integer? [#permalink] New post   15 Jun 2022, 16:00
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Avish11 wrote:

Hey, I reckon, based on your explanation, you have mistaken the first statement as x(squared) is a multiple of 2, instead of sqrt(x) is a multiple of 2

My answer is A

Sqrt (x) is a multiple of 2 so if sqrt(x) is anything 0,2,4,6,8,10, its square will also be divisible by 2 hence an even integer

Statement 2 is insufficient
X(squared) can be 4 (divisible by 4) which means x is 2 and YES EVEN
But if x squared is 12 (divisible by 4), x is sqrt(12), not an even number


You're absolutely right!
I somehow misread statement 1 as "\(x^2\) is a multiple of 2"

I have edited my response accordingly.
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Re: Is x an even integer? [#permalink] New post   16 Jun 2022, 08:05
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I thought I can add some value here in terms of building a process to solve such questions.

TBH, These types of questions can often catch someone off guard. So how do we make sure that we do not commit any mistakes?

    We always suggest that you form equations and then infer from the same.


Statement 1:
    \(\sqrt{x}\) is a multiple of 2.
    So we write them as, \(\sqrt{x}\) = 2K
    Hence, x = 4\(k^2\)

It is very easy to infer from this that x must be a multiple of 2(actually a multiple of 4). We eliminate options B, C, and E.

Statement 2
    \(x^2\) is a multiple of 4
    So we write, \(x^2\)=4p
    Or, x =2\(\sqrt{p}\) or -2\(\sqrt{p}\)

So it will be an even integer when p is a perfect square and if p is not then it will be an irrational number.
So this is not sufficient and we eliminate option D.

The answer is A
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Re: Is x an even integer? [#permalink] New post   23 Jun 2022, 09:08
Hello experts. Just one question, can zero not satisfy both the statements above?

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Re: Is x an even integer? [#permalink] New post   23 Jun 2022, 09:43
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Shikhar22 wrote:
Hello experts. Just one question, can zero not satisfy both the statements above?

Yes, and x = \(\sqrt{8}\) (for example) also satisfies (2). So based on (2) we cannot say for sure whether x is an even integer. Hence, (2) is insufficient.
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Re: Is x an even integer? [#permalink] New post   23 Jun 2022, 10:15
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This is the kind of Data Sufficiency that you need to decide if you go for algebra or plug in numbers.
I decided for both.
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Re: Is x an even integer? [#permalink] New post   27 Nov 2023, 11:09
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Re: Is x an even integer? [#permalink]
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