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Is x an even integer?

Bunuel

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28 Apr 2015, 05:15

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Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

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ak1802

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Updated on: 28 Apr 2015, 07:25

Originally posted by ak1802 on 28 Apr 2015, 06:55.
Last edited by ak1802 on 28 Apr 2015, 07:25, edited 1 time in total.

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Do St(1) and St(2) imply x^2/4 and x^4/16 have integer solutions? Since this is 700 level I'm just getting caught up on the wording of the problem.

Either way, if St(1) and St(2) aren't constrained by integer solutions the answer is E. The number 1 can be used to disprove each solution.

Additionally the square root of 8 squared is divisible by 4 in St(1) which makes X a non-integer. The square root of 8^4 is also divisible by 16, however x is still a non integer.

Answer choice - E.

Bunuel

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28 Apr 2015, 07:05

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ak1802

Do St(1) and St(2) imply x^2/4 and x^4/16 have integer solutions? Since this is 700 level I'm just getting caught up on the wording of the problem.

Either way, if St(1) and St(2) aren't constrained by integer solutions the answer is E. The number 1 can be used to disprove each solution. Additionally the square root of 8 squared is divisible by 4 in St(1) which makes X a non-integer. The square root of 8^4 is also divisible by 16, however x is still a non integer.

Answer choice - E.

x^2 is divisible by 4 means:
1. x^2 is an integer;
2. x^2/4 is an integer.

Generally, on the GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is a multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that:
(i) \(a\) is an integer;
(ii) \(b\) is an integer;
(iii) \(\frac{a}{b}=integer\).

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Updated on: 28 Apr 2015, 07:32

Originally posted by bsattin1 on 28 Apr 2015, 07:27.
Last edited by bsattin1 on 28 Apr 2015, 07:32, edited 1 time in total.

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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

ak1802

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28 Apr 2015, 07:31

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Bunuel

ak1802

Do St(1) and St(2) imply x^2/4 and x^4/16 have integer solutions? Since this is 700 level I'm just getting caught up on the wording of the problem.

Either way, if St(1) and St(2) aren't constrained by integer solutions the answer is E. The number 1 can be used to disprove each solution. Additionally the square root of 8 squared is divisible by 4 in St(1) which makes X a non-integer. The square root of 8^4 is also divisible by 16, however x is still a non integer.

Answer choice - E.

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.

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28 Apr 2015, 07:34

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Bunuel

Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Kudos for a correct solution.

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

ak1802

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28 Apr 2015, 07:52

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bsattin1

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.

Bunuel

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04 May 2015, 04:45

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Bunuel

Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Kudos for a correct solution.

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.

akara2500

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Bunuel

Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Kudos for a correct solution.

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

Bunuel

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akara2500

Bunuel

Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Kudos for a correct solution.

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.

hiteshlimbasiya

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05 Jun 2025, 09:27

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[ltr]Each statement alone is sufficient to determine if x is an even integer.
Here's why:
Statement (1): x2 is divisible by 4.
If x is an integer:

If x is even, then x can be written as 2k for some integer k. So, x2=(2k)2=4k2. This is clearly divisible by 4.
If x is odd, then x can be written as 2k+1 for some integer k. So, x2=(2k+1)2=4k2+4k+1=4(k2+k)+1. This leaves a remainder of 1 when divided by 4, so it's not divisible by 4.

Since statement (1) says x2 is divisible by 4, x must be an even integer.
Thus, statement (1) alone is sufficient.

[hr]
Statement (2): x4 is divisible by 16.
If x is an integer:

If x is even, then x can be written as 2k for some integer k. So, x4=(2k)4=16k4. This is clearly divisible by 16.
If x is odd, then x can be written as 2k+1 for some integer k. An odd number raised to any positive integer power is always odd. Therefore, x4 would be an odd number. An odd number cannot be divisible by 16 (an even number). For example, if x=1, x4=1, not divisible by 16. If x=3, x4=81, not divisible by 16.

Since statement (2) says x4 is divisible by 16, x must be an even integer.
Thus, statement (2) alone is sufficient.
Since each statement alone is sufficient to determine that x is an even integer, the answer is that either statement alone is sufficient.[/ltr]

hiteshlimbasiya

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[ltr]Each statement alone is sufficient to determine if x is an even integer.
Here's why:
Statement (1): x2 is divisible by 4.
If x is an integer:

If x is even, then x can be written as 2k for some integer k. So, x2=(2k)2=4k2. This is clearly divisible by 4.
If x is odd, then x can be written as 2k+1 for some integer k. So, x2=(2k+1)2=4k2+4k+1=4(k2+k)+1. This leaves a remainder of 1 when divided by 4, so it's not divisible by 4.

Since statement (1) says x2 is divisible by 4, x must be an even integer.
Thus, statement (1) alone is sufficient.

[hr]
Statement (2): x4 is divisible by 16.
If x is an integer:

If x is even, then x can be written as 2k for some integer k. So, x4=(2k)4=16k4. This is clearly divisible by 16.
If x is odd, then x can be written as 2k+1 for some integer k. An odd number raised to any positive integer power is always odd. Therefore, x4 would be an odd number. An odd number cannot be divisible by 16 (an even number). For example, if x=1, x4=1, not divisible by 16. If x=3, x4=81, not divisible by 16.