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# Is a even?

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Updated on: 16 Jul 2013, 00:06
2
1
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Difficulty:

5% (low)

Question Stats:

83% (00:24) correct 17% (00:22) wrong based on 405 sessions

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Is a even?

(1) 2a is even.
(2) $$\sqrt{a}$$ is even.

Originally posted by cyberjadugar on 24 May 2012, 03:38.
Last edited by Bunuel on 16 Jul 2013, 00:06, edited 2 times in total.
Edited the OA.
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24 May 2012, 04:16
3
2
Hi,

My mistake, option two needs to be corrected as $$a^2$$, in that case the answer is E.

Regards,

No, in this case the answer would be C.

Is a even?

(1) 2a is even --> $$2a=even$$ --> $$a=\frac{even}{2}$$, so $$a$$ is either even or odd (in any case $$a$$ must be an integer). Not sufficient.

(2) $$a^2$$ is even --> $$a^2=even$$ --> $$a=\sqrt{even}$$, so $$a$$ is either even (for example if $$a^2=4$$) or an irrational number (for example if $$a^2=6$$). Not sufficient. Generally $$a^2$$ to be an integer $$a$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (notice that $$a$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$a^2$$ won't be an integer).

(1)+(2) Since from (1) $$a$$ is an integer, then in order $$a^2$$ to be even $$a$$ must be even too. Sufficient.

Hope it's clear.
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24 May 2012, 03:44
Is a even?

(1) 2a is even --> a can be even as well as odd. Not sufficient.

(2) $$\sqrt{a}$$ is even --> $$\sqrt{a}=even$$ --> $$a=even^2=even$$. Sufficient.

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24 May 2012, 03:58
1
Hi,

My mistake, option two needs to be corrected as $$a^2$$, in that case the answer is E.

Regards,
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24 May 2012, 05:18
Thanks for the clarification!
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24 May 2012, 09:03
for the incorrect problem picked B. great explanation by bunuel of the original problem. especially liked the representation of expression such as even/2. makes analysis much simpler.
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12 Jun 2013, 04:27
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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12 Jun 2013, 23:05
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Is a even?

(1) 2a is even.
(2) $$\sqrt{a}$$ is even.

(1) we need to remember that when whatever (even odd numbers) numbers are multiplied by 2 the result will be an even number. In the first statement a could be odd, so the statement is not sufficient;

(2) if $$\sqrt{a}$$ is even that means that a consists of at least two identical factors and that factor is even. Since that two identical factors are similar then their product (in this case it is a) must be even as well. So this statement is sufficient. Answer is B
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04 May 2014, 04:46
Hi Bunnel

i didnt understand the explanation for st 2 in revised qs.

(2) a^2 is even --> a^2=even --> a=\sqrt{even}, so a is either even (for example if a^2=4) or an irrational number (for example if a^2=6). Sufficient. Generally a^2 to be an integer a must be either an integer or an irrational number (for example: \sqrt{3}), (notice that a can not be reduced fraction, for example \frac{2}{3} or \frac{11}{3} as in this case a^2 won't be an integer).

the qs is asking if a is even... if a^2 = 6... then also a is even right? as the qs doesnt mention integer anywhere.
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04 May 2014, 05:04
nandinigaur wrote:
Hi Bunnel

i didnt understand the explanation for st 2 in revised qs.

(2) a^2 is even --> a^2=even --> a=\sqrt{even}, so a is either even (for example if a^2=4) or an irrational number (for example if a^2=6). Not sufficient. Generally a^2 to be an integer a must be either an integer or an irrational number (for example: \sqrt{3}), (notice that a can not be reduced fraction, for example \frac{2}{3} or \frac{11}{3} as in this case a^2 won't be an integer).

the qs is asking if a is even... if a^2 = 6... then also a is even right? as the qs doesnt mention integer anywhere.

Don't follow you... If $$a^2=6$$, then $$a=\sqrt{6}\neq{integer}$$.
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04 May 2014, 05:19
i mean to say that a=sqrt 6 is even right? how do we know that it has to be an interger.
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04 May 2014, 05:29
oh i get it... we can classify only integers as odd or even... irrational numbers cannot be even or odd.. so when we say a sqrt6... is it an irrational number which cannot be even or odd. am i right?

"key take away": only integers can be odd or even right?
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04 May 2014, 06:06
nandinigaur wrote:
oh i get it... we can classify only integers as odd or even... irrational numbers cannot be even or odd.. so when we say a sqrt6... is it an irrational number which cannot be even or odd. am i right?

"key take away": only integers can be odd or even right?

Yes, only integers can be even or odd:

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.

An odd number is an integer that is not evenly divisible by 2.

Theory on Number Properties: math-number-theory-88376.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

Hope this helps.
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19 Jan 2018, 10:32
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is a even? &nbs [#permalink] 19 Jan 2018, 10:32
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