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by sana.noor » Sun May 05, 2013 9:55 am
If x is an integer, is (x2 - 1)(x - 3) an even number?
(1) x is even.
(2) Each prime factor of (x2 + 1) is greater than 3.

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Source: — Data Sufficiency |

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by killerdrummer » Sun May 05, 2013 10:26 am
Given : X is Integer
(1) x is even.
(x2 - 1)(x - 3) an even number
Square of any even number is even and when we subtract any odd from even you get odd.

So, (x2 - 1) is odd

(x - 3) is also odd as Even-Odd =Even

As both the numbers are odd.Therefore their product will always be Odd.


Sufficient
(2) Each prime factor of (x2 + 1) is greater than 3.
In other words X is even same as above.

How??? let us see

Given : Each prime factor of (x2 + 1) is greater than 3

Means 2 can not be prime factors or in other words (x2 + 1) has to be odd.

and this is possible only when x is even as we know square of odd is odd and adding 1 to it makes (x2 + 1)even.

Therefore x has to be even.

Example:
plug 2 in x2 + 1 = 5 Correct
plug 4 in x2 + 1 = 17 Correct
Plug 3 in x2 + 1 = 10 Incorrect as 2 is prime factor of 10.


As x is even in this case also.As shown above this is sufficient too!!


Correct answer D
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by Blue_Skies » Tue May 07, 2013 12:07 pm
If x is an integer, is (x2 - 1)(x - 3) an even number?
(1) x is even.
(2) Each prime factor of (x2 + 1) is greater than 3.

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the product of two integers are even if at least one of them is even.

From 1 we know the product is odd. Sufficient.
From 2 We know that all the factors of 2x+1 are odd. This is only possible if the number itself is odd. Hence x is even. Sufficient

D