If x is a positive integer, is x odd?

Question

(1) x^3 + x^2 is an even integer.
(2)  \sqrt{x}+1  is an odd integer.

chetan2u · Answer

We know x is positive, and we have to find if x is ODD..

lets see the staements--

(1) x^3 + x^2 is an even integer.
For EVEn either both terms are ODD or both EVEN..
Insuff

(2)  \sqrt{x}+1  is an odd integer
Clearly  \sqrt{x}  is EVEN...
so x will also be EVEN...
Suff

B

Youraisemeup · Answer

A is insuff. as x^3 and x^2 could be both even or both odd..

B is suff. as root x+1=odd
thus root x=even
and x=even.

stonecold · Answer

Here B is sufficient as √x and x will have the same even odd nature

MathRevolution · Answer

If x is a positive integer, is x odd?

(1) x^3 + x^2 is an even integer.
(2) x √ +1 x+1 is an odd integer. 
 
There is 1 variable (x) in the original condition. In order to match the number of variables to the number of equations, we need 1 more equation. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that D is the correct answer choice.
In case of the condition 1), if x^3 + x^2 =even, the answer becomes yes when x=1 and the answer becomes no when x=2. Therefore, the condition is not sufficient.
In case of the condition 2), from Root(x)+1=odd, we can obtain Root(x)=odd+1=even. Then we get x=even^2=even. The answer is yes and the condition is sufficient. Therefore, the correct answer choice is B.

l 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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If x is a positive integer, is x odd?

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If x is a positive integer, is x odd?

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