Is a even? : GMAT Data Sufficiency (DS)
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Is a even?

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Is a even? [#permalink]

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Is a even?

(1) 2a is even.
(2) \(\sqrt{a}\) is even.
[Reveal] Spoiler: OA

Last edited by Bunuel on 15 Jul 2013, 23:06, edited 2 times in total.
Edited the OA.
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Re: Is a even? [#permalink]

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cyberjadugar wrote:
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.

Answer: C.

Hope it's clear.
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Re: Is a even? [#permalink]

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New post 12 Jun 2013, 22:05
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cyberjadugar wrote:
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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Re: Is a even? [#permalink]

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Re: Is a even? [#permalink]

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New post 24 May 2012, 02:58
Hi,

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

Regards,
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Re: Is a even? [#permalink]

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New post 24 May 2012, 04:18
Thanks for the clarification! :-D
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Re: Is a even? [#permalink]

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New post 24 May 2012, 08:03
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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New post 12 Jun 2013, 03:27
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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New post 04 May 2014, 03:46
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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Re: Is a even? [#permalink]

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New post 04 May 2014, 04:04
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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New post 04 May 2014, 04:19
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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New post 04 May 2014, 04:29
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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Re: Is a even? [#permalink]

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New post 04 May 2014, 05:06
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
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Hope this helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is a even?   [#permalink] 04 May 2014, 05:06
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