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Is z an even integer?

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Is z an even integer? [#permalink]

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Is z an even integer?

(1) \(z^2\) is an even integer.

(2) y*z is an even integer, where z and y are integers.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Apr 2017, 02:37, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is z an even integer? [#permalink]

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New post 29 Apr 2017, 08:13
Statement 1 alone is sufficient and statement 2 is not sufficient.


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Re: Is z an even integer? [#permalink]

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Munch wrote:
Statement 1 alone is sufficient and statement 2 is not sufficient.


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Hi,
What if z is √2...
\(z^2=2\) and thus an even integer but z is not an even integer.
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: Is z an even integer? [#permalink]

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New post 29 Apr 2017, 14:03
chetan2u wrote:
Is z an even integer?
1) \(z^2\) is an even integer.
2) y*z is an even integer, where z and y are integers.



from 1

z could be a radical ... insuff

from 2

if y is even integer and no idea about z... insuff

both

z is an integer and z^2 is even thus z is even...C
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Re: Is z an even integer? [#permalink]

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New post 29 Apr 2017, 21:37
Statement 1: Z can be sqrt2 insufficient
Statement 2 : Not Unique answer.

Both we can derive unique answer that Z is an integer and even
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Re: Is z an even integer? [#permalink]

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New post 29 Apr 2017, 22:09
Even/Odd Questions are one of my favourite.
So much to learn.

Here is what i did on this one ->

Firstly => The most important step in any even/odd question before we move to the individual statements is check whether the involved numbers are integers or not.

We are not told that z is an integer.

Lets roll-->


Statement 1 =>
z^2 is even

If z is an integer => z would be even.
But what if z=\(√12\) ?
z^2 is even
But \(√12\) isnt even.

Not sufficient.

Statement 2 =>
So y and z are integers.

yz= even
So at-least one out of y or z must be even.
So if y is even => z can be even or odd.

Not sufficient.

Combing the two statements => As z is an integer and z^2 is even => z must be even.
Property in action =>Positive exponent does not affect the even/odd nature of any integer.

Hence C is sufficient.


SMASH THAT C.

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Re: Is z an even integer? [#permalink]

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New post 30 Apr 2017, 03:12
chetan2u wrote:
Is z an even integer?

(1) \(z^2\) is an even integer.

(2) y*z is an even integer, where z and y are integers.


(1) \(z^2\) is an even integer.

\(z^2=2 \implies z=\sqrt{2}\) is not an even integer.
\(z^2=4 \implies z=2\) is an even integer.

Hence (1) is insufficient.

(2) \(y\) and \(z\) are both integers.

Since \(y \times z\) is even, \(y\) or \(z\) or both must be even.

If \(y\) is odd, then \(z\) must be even.
If \(y\) is even, then \(z\) could be odd.

Hence, insufficient.

Cobine (1) and (2), we have \(z\) is an integer and \(z^2\) is even. Hence \(z\) must be even. Sufficient.

The answer is C.
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Re: Is z an even integer? [#permalink]

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New post 10 Dec 2017, 14:17
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Hi All,

We're asked if Z is an EVEN INTEGER. This is a YES/NO question. We can solve it by TESTing VALUES.

1) Z^2 is an EVEN integer.

IF...
Z^2 = 2, then Z = (root2) and the answer to the question is NO.
Z^2 = 4, then Z = 2 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) (Y)(Z) is an EVEN integer and Z and Y are INTEGERS

IF...
Y = 1 and Z = 2, then the answer to the question is YES.
Y = 2 and Z = 1, then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know...
Y and Z are INTEGERS
Z^2 is an EVEN integer

The ONLY way to square an INTEGER and end up with an EVEN integer is if you started with an EVEN integer. Fact 2 tells us that Z MUST be an integer, so Z MUST be even and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT

Final Answer:
[Reveal] Spoiler:
C


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Re: Is z an even integer?   [#permalink] 10 Dec 2017, 14:17
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