M04-07

Question

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.

Bunuel · Accepted Answer

Official Solution:

If  x  is an integer, what is the value of  x ? 
 
(1)  |23x|  is a prime number. 
 
Since 23 is a prime number, this statement implies that  x=1  or  x=-1 . Not sufficient.
 
(2)  2\sqrt{x^2}  is a prime number. 
 
We can rewrite the expression as  2\sqrt{x^2}=2|x| . Since 2 is a prime number, this statement implies that  x=1  or  x=-1 . Not sufficient.
 
(1)+(2) Using both conditions,  x  could be either 1 or -1. The information provided is still insufficient to determine the exact value of  x .

Answer: E

Gmatdecoder · Answer

How can a negative number be prime? Statement II should limit value to x=1 for prime number 2, I think.. please explain

Bunuel · Answer

Yes, only positive numbers can be primes but if you look closely to the second statement you should notice that even for x=-1 you get POSITIVE result: 2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1^2}=2*1=2=prime .

Avinash2303 · Answer

Bunuel

Since \sqrt{x} = lxl i.e. x can be +ve as well as -ve, can you please make me understand why \sqrt{(-1) (-1)} is not equals to (-1). We can take one negative value out of root, and multiple of 2*(-1) would be negative i.e. non-prime number.

Why we have to calculate \sqrt{(-1) (-1)} before taking value out of the root? I know this way it would be 2 and a prime number.

Bunuel · Answer

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root.   That is:

\sqrt{9} = 3 , NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;

Notice that in contrast, the equation  x^2 = 9  has TWO solutions, +3 and -3. Because  x^2 = 9  means that  x =-\sqrt{9}=-3  or  x=\sqrt{9}=3 .

ameralhajj · Answer

why is it wrong to simplify the second statement to 2*x since the sq.root and the square cancel each other? If i do that then for the second statement to be true, only when x=1 will the result be a prime.

Bunuel · Answer

Because  \sqrt{x^2}=|x| , not x. Try plugging in a negative numbers to check.

Why does  \sqrt{x^2}=|x| ?

The point here is that  square root function cannot give negative result : wich means that  \sqrt{some \ expression}\geq{0} .

So  \sqrt{x^2}\geq{0} . But what does  \sqrt{x^2}  equal to?

Let's consider following examples:
If  x=5  -->  \sqrt{x^2}=\sqrt{25}=5=x=positive ;
If  x=-5  -->  \sqrt{x^2}=\sqrt{25}=5=-x=positive .

So we got that:
 \sqrt{x^2}=x , if  x\geq{0} ; 
 \sqrt{x^2}=-x , if  x<0 .

What function does exactly the same thing? The absolute value function! That is why  \sqrt{x^2}=|x|

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

M04-07

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