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# If x is an integer, what is the value of x?

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If x is an integer, what is the value of x? [#permalink]

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01 Jul 2013, 19:54
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51% (01:43) correct 49% (00:38) wrong based on 35 sessions

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If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) $$2*\sqrt{x^2}$$ is a prime number

M04-07

[Reveal] Spoiler:
Now, I get why statement 1 is not sufficient (x could be 1 or -1). But the OA says that in statement 2, x could still be 1 or -1. Is it not the rule that the square root of a number only yields a positive number. Shouldn't x be here. Help.
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Jul 2013, 22:09, edited 1 time in total.
Renamed the topic and edited the question.
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01 Jul 2013, 20:55

Stmt 1) Not sufficient since x can be only -1 or 1 because |23x| = 23x when x > 0 and -23x when x < 0

Stmt 2) Not sufficient since x can be only -1 or 1 because sqrt(x*2) = |x|

Considering Stmt 1) and Stmt 2) x can be -1 or 1
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01 Jul 2013, 21:42
srujani wrote:
If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) $$2*\sqrt{x^2}$$ is a prime number

What you are saying is exactly correct : Square root of a number will always be a non-negative entity. The question doesn't wrestle that point.
$$\sqrt{x^2} = |x| \geq{0}.$$. However, as you can see, both x = 1 and x = -1 are valid as because |1| = |-1| = 1.

So basically, $$(-1)^2 = (1)^2 = 1$$, but $$\sqrt{1}$$ = 1.

From F. S 1, we know that |23x| is a prime number for both x =1 and x=-1. Thus,no unique value of x is present. Insufficient

From F.S 2, $$2*\sqrt{x^2}$$ can be a prime no only if $$x^2 = 1$$. Again, $$x^2 = 1$$ $$\to$$ $$x = \pm1$$. Just as above, we get 2 values of x, hence Insufficient.

Even after combining both the fact statements, no unique value of x can be found.

E.
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Re: If x is an integer, what is the value of x? [#permalink]

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01 Jul 2013, 22:12
srujani wrote:
If x is an integer, what is the value of x?

(1) |23x| is a prime number
(2) $$2*\sqrt{x^2}$$ is a prime number

M04-07

[Reveal] Spoiler:
Now, I get why statement 1 is not sufficient (x could be 1 or -1). But the OA says that in statement 2, x could still be 1 or -1. Is it not the rule that the square root of a number only yields a positive number. Shouldn't x be here. Help.

1. If x is an integer, what is the value of x?

(1) $$|23x|$$ is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) $$2\sqrt{x^2}$$ is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Discussed here: new-set-number-properties-149775-40.html#p1205341

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Re: If x is an integer, what is the value of x? [#permalink]

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02 Jul 2013, 22:12
Bunuel,

As per the explanation for the option 2) , x can be either +1 or -1. But since 2 sqrt(x^2) is a prime number, can we not assume that x can't be -1 because a prime number is "an integer that has no integral factors but itself and 1" which will not be the case if x is -1.
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Re: If x is an integer, what is the value of x? [#permalink]

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02 Jul 2013, 22:20
kinghyts wrote:
Bunuel,

As per the explanation for the option 2) , x can be either +1 or -1. But since 2 sqrt(x^2) is a prime number, can we not assume that x can't be -1 because a prime number is "an integer that has no integral factors but itself and 1" which will not be the case if x is -1.

If you plug $$x=-1$$ into $$2\sqrt{x^2}$$ you'll still get a prime number:

$$2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2=prime$$.
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Re: If x is an integer, what is the value of x?   [#permalink] 02 Jul 2013, 22:20
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