If x is an integer and x^2>1, is x a positive integer?

Given, \(x^2>1\) implies that |x|>1
Or, x < -1, x >1 (x: all the integers except -1,0,1)----------(a)

Question stem :- Is x>0 ?

St1:- \(x^2=6x\)
Or, \(x^2-6x=0\)
Or, x(x-6)=0
So, x=6,0--------------(b)
From (a)&(b), x=6.
Sufficient.

St2:- \(x=|y|\), where y is an integer
Here |y| is an absolute expression, so x is always positive irrespective of sign of y. However, y may take +ve or -ve polarity.
Hence, x is positive.
Sufficient.
NOTE:- We have to discard the following possible values of y hence x.
1) y=0,1,-1; This as per our observation obtained at (a).

Ans. (D)

[quote="Afc0892"]From statement 1:

X^2-6X = 0
X(X-6) = 0
X = 0 or 6.

Since X^2>0 then X = 6.
1 is sufficient.

From statement 2: X = +Y or -Y.

No info about Y.
Hence insufficient.

A is the answer.

Hey, In statement 2, x = |y|. u don't have to think about the value of y or it's sign. Regardless of y's sing , x is always positive.

Thanks.

so it shud b "D" know?

(1) \(x^2=6x\)
This statement is sufficient to tell us that x is +ve

(2) \(x=|y|\), where y is an integer
|y|= + y = x= x is a positive integer
sufficient

D

Hi Chetan, in my Evaluation E should be the answer. Could you please specify if my evaluation is wrong.

Given in question: x-->Integer
x^2>1

Now x^2>1 can be solved further as:-

(x^2-1)>0
(x+1)(x-1)>0
this gives us two possibilities:-

x+1>0 & x-1>0
x>-1 & x>1

So this implies that x = 0,1,2,3,4.....


Statement-1:-
x^2=6x
this can be further simplified to
x^2-6x=0
x(x-6)=0
this gives us two possible answers
x=0 or x=6
hence, x can be positive or Zero.
So, Statement-1 is not sufficient


Statement-2:-
x=IyI (I I-->Modulus)
given that y --> Integer
So, y = -2,-1,0,1,2,......
if y = 0 then x = 0
if y = -2,-1,1,2... then x = 2,1,1,2...
Hence, x can be positive or zero.
So, Statement-2 is not sufficient


Together Statement-1 & 2:-
By combining the two statements also we have two possible scenarios of x
x can be positive or Zero.
Hence, not sufficient.

So Answer: E

We are given x^2>1 in the question, so x<-1 or X>1 but X cannot be 0..
If x=0, then x^2=0 but we are given x^2>1

Hence only non zero value is possible...
Each statement gives us a definite answer

Asked: If x is an integer and \(x^2>1\), is x a positive integer?


(1) \(x^2=6x\)
x^2=6x>1; x>1/6>0; Since x is an integer & x>0; x is a positive integer
SUFFICIENT

(2) \(x=|y|\), where y is an integer
x = |y| > 0
Since x is an integer & x>0; x is a positive integer
SUFFICIENT

IMO D