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

Question

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

M04-07

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.

Bunuel · Accepted Answer

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. Answer: E. Discussed here: new-set-number-properties-149775-40.html#p1205341 P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html#p1092822 (Posting rules. Pay attention to the rule #3.) rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Formatting: highlight the whole expression and press m button.)

nt2010 · Answer

IMO answer is E

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

mau5 · Answer

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   	o   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.

kinghyts · Answer

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.

Bunuel · Answer

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 .

sampad · Answer

i dont agree with the solution. if mod of 23x is a prime number then as per the definition of prime number 23x is always positive. So how will x have two values?

and as per the statement 2, sqrt of (x2) is always positive. and when it was mentioned it should be prime then how we can take the value of -1.

i dont agree with the solution. There are many problems i did where we took positive value when square root was given.

Bunuel · Answer

The solutions above are correct.

For (1):
If x = 1, then |23x| = |23*1| = |23| = 23 = prime.
If x = -1, then |23x| = |23*(-1)| = |-23| = 23 = prime.

For (2):
If x = 1, then  2\sqrt{x^2}=2\sqrt{1}=2=prime .
If x = -1, then  2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1}=2=prime .

Hope it's clear now.

Sukriti10 · Answer

Hi Bunuel!
The question is asking whether x is an integer or not, does the value of x really matters then? 
I understand that after solving both statements we get x = 1 or -1
However, both 1 and -1 are integers. 
Pls tell if there needs to be unique value of x in order for us to reach an answer?
Thanks!

Bunuel · Answer

Here is the question: If x is an integer,  what is the value of x?

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

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