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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.

(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 _________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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.

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. _________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

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 \(a=\sqrt{6}\neq{integer}\). _________________

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? _________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

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.

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...