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M04-07

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Bunuel
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M04-07 [#permalink] New post   16 Sep 2014, 00:22
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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.
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Re M04-07 [#permalink] New post   16 Sep 2014, 00:22
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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
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Re: M28-48 [#permalink] New post   28 Oct 2015, 01:54
How can a negative number be prime? Statement II should limit value to x=1 for prime number 2, I think.. please explain
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Re: M28-48 [#permalink] New post   28 Oct 2015, 06:16
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Gmatdecoder wrote:
How can a negative number be prime? Statement II should limit value to x=1 for prime number 2, I think.. please explain


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:\(2\sqrt{x^2}=2\sqrt{(-1)^2}=2\sqrt{1^2}=2*1=2=prime\).
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Re: M28-48 [#permalink] New post   09 Jul 2018, 17:04
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.
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Re: M28-48 [#permalink] New post   09 Jul 2018, 20:28
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Avinash2303 wrote:
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.


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;
\(\sqrt[4]{16} = 2\), 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 \(x=\sqrt{9}=3\).
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Re: M28-48 [#permalink] New post   08 Jan 2019, 07:08
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. :(
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Re: M28-48 [#permalink] New post   08 Jan 2019, 07:40
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ameralhajj wrote:
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. :(


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 \(\sqrt{x^2}=|x|\)
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Re: M28-48 [#permalink] New post   22 Apr 2023, 11:09
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M04-07 [#permalink] New post   12 Aug 2023, 01:58
I think this is a high-quality question and I agree with explanation.
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Re M04-07 [#permalink]
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