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If x is an integer, is x^3 even?

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If x is an integer, is x^3 even? [#permalink]

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New post 01 Apr 2016, 02:25
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If x is an integer, is x^3 even? [#permalink]

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New post 01 Apr 2016, 02:51
Bunuel wrote:
If x is an integer, is \(x^3\) even?

(1) 2x + 2 is even.

(2) 3x + 1 is even.


Given: x is an integer
Required: Is \(x^3\) even?
Or in other words, is x even?

Statement 1: 2x + 2 is even
Since x is multiplied by an even number, it will always be even irrespective of the value of x.
This is added to 2, which is also an even number.
Hence 2x + 2 will always be even. Nothing can be said about x
INSUFFICIENT

Statement 2: 3x + 1 is even.
Here 1 is odd, therefore for 3x + 1 to be even, 3x has to be odd.
Now 3 is odd, hence x has to be odd for 3x to be odd.
SUFFICIENT

Option B

NOTE:
Even + Even = Even
Odd + Even = Odd
Odd + Odd = Even

Even * Even = Even
Odd * Even = Even
Odd * Odd = Odd

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If x is an integer, is x^3 even? [#permalink]

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New post 02 Apr 2016, 11:32
If x is an integer, is x^3 even?
x^3 will be even when x is even. Basically we need to find whether x is odd or even.

(1) 2x + 2 is even.
Both odd and even values of x satisfies the above eq. Insufficient.

(2) 3x + 1 is even.
it can b even only for odd values of x.
so the answer for the statement " x^3 is even" is NO. Sufficient.

hence B.

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Re: If x is an integer, is x^3 even? [#permalink]

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New post 03 Apr 2016, 03:25
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x is an integer, is x^3 even?

(1) 2x + 2 is even.

(2) 3x + 1 is even.


In the original condition, x^3=even? becomes x=even?. Then, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), 2x+2=even, 2x=even-2=even -> x=1 no, x=2 yes, which is not sufficient.
For 2), 3x+1=even, 3x=even-1=odd -> x=odd, which is no and sufficient.
Therefore, the answer is B.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x is an integer, is x^3 even? [#permalink]

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New post 07 Apr 2016, 09:14
statement A--clearly insufficient had x been even or odd.
statement B means, 3x+1 is even, which means 3x is odd. This can only happen only if x is also odd.

Thus Sufficient.

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Re: If x is an integer, is x^3 even? [#permalink]

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New post 25 Sep 2017, 19:45
Bunuel wrote:
If x is an integer, is x^3 even?

(1) 2x + 2 is even.

(2) 3x + 1 is even.


The question is really...is X odd?

Statement 1

x could be odd or even

insuff

Statement 2

x must be odd

suff

B

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Re: If x is an integer, is x^3 even?   [#permalink] 25 Sep 2017, 19:45
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