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Bunuel
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I agree with rahul16singh28.

In statement 1 is said x^2 is divisible by 4, x^2= a*4 ---> x=a^(1/2)*2 (not considering sings)

so, a could be 1.5 and x= 3 (not an even integer) , but a could be 9 and x= 6 (even integer). Not sufficient

Statement two is no different than statement 1. Also not sufficient

And both together Not sufficient
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Bunuel
Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.
I think E is the answer as x could be 4root 3 or 4 we can't saw anything.
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The trick in this question is that it wants us to think of x in terms ox multiples of 2, such as 2, 4, 8 ... If you plug in any of these numbers, for sure they are divible by 4 (in statement I) or by 16 (in statement II)

We should ask ourselves if there is a non integer number x that when x^2 is divisible by 4 or when x^4 is divisible by 16.

(2√2)^2 is equal to 8 and 8 is divisible by 4
(2 * 2^1/4) ^ 4 is equal to 32 and 32 is divisible by for.

Together, (2 * 2^1/4) ^ 2 is not divisible by 4, so there is at least one number that is sufficient for statements I and II but is not an integer.

Answer: E
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rahul16singh28
Bunuel
Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Statement I:

Let \(x = \sqrt{20}, x^2\) is Divisible by 4.
Let \(x = 2, x^2\) is Divisible by 4.........So, Insufficient.

Statement II:

\(x = 2(2)^{1/4}, x^4\) is Divisible by 16.
Let, \(x= 2, x^4\) is Divisible by 16......So, Insufficient.

Combining I & II:

Let \(x = \sqrt{20},\)
\(x^2\) is Divisible by 4 and \(x^4\) is Divisible by 16.

Let \(x = 2\)
\(x^2\) is Divisible by 4 and \(x^4\) is Divisible by 16.

So, Insufficient. Hence, E.
Would u explain highlighted part ,plz?
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vivapopo
rahul16singh28
Bunuel
Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Statement I:

Let \(x = \sqrt{20}, x^2\) is Divisible by 4.
Let \(x = 2, x^2\) is Divisible by 4.........So, Insufficient.

Statement II:

\(x = 2(2)^{1/4}, x^4\) is Divisible by 16.
Let, \(x= 2, x^4\) is Divisible by 16......So, Insufficient.

Combining I & II:

Let \(x = \sqrt{20},\)
\(x^2\) is Divisible by 4 and \(x^4\) is Divisible by 16.

Let \(x = 2\)
\(x^2\) is Divisible by 4 and \(x^4\) is Divisible by 16.

So, Insufficient. Hence, E.
Would u explain highlighted part ,plz?

I think what he means is that
Statement II:
Let, \(x = 2*2^\frac{1}{4}\) (not an even integer) --> \(x^4 = 2^4 * 2 = 16*2\) --> \(x^4\) is divisible by 16
Let, \(x= 2\) (an even integer), \(x^4\) is Divisible by 16
--> Insufficient.
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Bunuel
Is x an even integer?

(1) x^2 is divisible by 4.
(2) x^4 is divisible by 16.

Asked: Is x an even integer?

(1) x^2 is divisible by 4.
If x^2 = 8; x = 2root(2); x is NOT an even integer
But if x^2 = 4; x = 2; x is an even integer
NOT SUFFICIENT

(2) x^4 is divisible by 16.
If x^4 = 64; x = 2root(2); x is NOT an even integer
But if x^4 = 16; x = 2; x is an even integer
NOT SUFFICIENT

(1) + (2)
(1) x^2 is divisible by 4.
If x^2 = 8; x = 2root(2); x is NOT an even integer
But if x^2 = 4; x = 2; x is an even integer
(2) x^4 is divisible by 16.
If x^4 = 64; x = 2root(2); x is NOT an even integer
But if x^4 = 16; x = 2; x is an even integer
NOT SUFFICIENT

IMO E
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Statement 1: x^2 is divisible by 4
if x^2 = 4, then it is divisible by 4, x = 2 => so x is an integer
if x^2 = 8, then it is divisible by 4, x = 2^(1/2) => so x is not an integer

Statement 2: x^4 is divisible by 16
if x^4 = 16, then it is divisible by 16, x = 2 => so x is an integer
if x^4 = 32, then it is divisible by 16, x = 2^(1/2) => so x is not an integer

Combining both also gives same cases
So Option E is the answer
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