Is x an even integer? (1) square root of x is a mutiple of 2

Question

Is x an even integer?

(1) square root of x is a mutiple of 2
(2) x^2 is a multiple of 4 

Any suggestions on how to answer this?

yogachgolf · Answer

1. If x=16, sqrt of 16=4, 4/2=2. x is an even integer
 If x=4, sqrt of 4=2, 2/2=1. x is not an even integer

2. If x=2, 2^2=4, 4/4=1. x is not an even integer
 If x=4, 4^2=16,16/4=4. x is an even integer

1&2 combined, insuff.

E

pmenon · Answer

i would say D.

All multiples of 2 and 4 are even #s

shubhampandey · Answer

It is E for above Reason. I didn't catch it when I solved it but it make sense

Ant · Answer

I also chose D, but the correct answer is A! Here is the explanation given:

Statement 1 by itself is enough. Plug in: if x = 4, then the answer to the question is “yes.” If x = 36, then the answer is “yes” again. No matter what you plug in for x, the answer will always be “yes.” Statement 2, however, is not. Statement 2 by itself is not enough. If x squared is a multiple of 4, x could be 2, or it could be something such as the square root of 24. So you don’t know if x is an even integer or not.

Not sure if this helps anyone, but it doesn't help me. There has to be any easier to solve this?

quote="yogachgolf"]

Curious about the OA...

johnrb · Answer

The idea's something like this:

I. sqrt(x) = 2k, where k is an integer.
So x = 4k^2. k is an integer, so k^2 is an integer, so x is an integer. And x is divisible by 4, so it must be even. SUFF.

II. x^2= 4k, where k is an integer.
So x = 2sqrt(k). But even though k's an integer, the square root of k may not be. (Suppose x^2 is 12. Then II is true. But sqrt(12) is not even an integer, so it certainly can't be an even integer.) INSUFF.

A's our answer.

Ravshonbek · Answer

Ant. the OE is fine. it is clearly explained.

imagine the same here. x=sqrt12 then x^2=12. this is divisible by 4 right?
 
but x=sqrt12 which is not and integer . it is the same as sqrt24 that's what the OE trying to say to us.

ian7777 · Answer

Also, recall that even times even is always event; odd times odd is always odd.

1 is essentially saying that the square root of x is even. So x must also be even.

We could use the same logic for 2, but, as has been pointed out, we don't know if it's an integer or not. You always have to think about that, particularly when the question asks specifically if x is an integer or not. 

Hope that adds some clarity.

pmenon · Answer

got A for this one. From statement 1, its pretty clear (to me anyways, for once

) that x will always be an even integer as root(x)=2,4,6,8,... squaring both sides will produce only integer values for x From statment 2, you can have x either as a integer, i.e. root(4)=2, or not, i.e. x=root(8)

Is x an even integer? (1) square root of x is a mutiple of 2

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