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# M28-48

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Math Expert
Joined: 02 Sep 2009
Posts: 47920

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16 Sep 2014, 01:44
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Difficulty:

55% (hard)

Question Stats:

50% (01:06) correct 50% (00:58) wrong based on 58 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.

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:44
Official Solution:

(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.

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Intern
Joined: 14 Oct 2015
Posts: 36
GMAT 1: 640 Q45 V33

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21 Oct 2015, 08:26
1
I think this is a high-quality question and I agree with explanation. This and M04-07 are duplicates
Intern
Joined: 23 Nov 2014
Posts: 32

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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
Math Expert
Joined: 02 Sep 2009
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28 Oct 2015, 06:16
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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Joined: 04 Sep 2016
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17 Apr 2017, 07:02
what about the rule : 'When the GMAT provides square root sign for an even root,such as x^(1/2) or x^(1/4), then the only accepted answer is the positive root''?
I am asking this because i used this rule to find only one solution (1) in the second statement. Is the rule only talking about x^(even power) ?
thanks
Math Expert
Joined: 02 Sep 2009
Posts: 47920

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17 Apr 2017, 07:23
alexlovesgmat wrote:
what about the rule : 'When the GMAT provides square root sign for an even root,such as x^(1/2) or x^(1/4), then the only accepted answer is the positive root''?
I am asking this because i used this rule to find only one solution (1) in the second statement. Is the rule only talking about x^(even power) ?
thanks

This rule is not violated. I tried to explain this here: https://gmatclub.com/forum/m28-184547.html#p1593296
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Joined: 29 May 2017
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14 Jun 2017, 09:29
I think this is a high-quality question and I don't agree with the explanation. Since a prime number can't be negative, why we are considering x = -1? this should always be x =1. isn't it?
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14 Jun 2017, 09:36
kutput wrote:
I think this is a high-quality question and I don't agree with the explanation. Since a prime number can't be negative, why we are considering x = -1? this should always be x =1. isn't it?

Yes, only positive integers can be primes but you get positive result with both 1 and -1. Please re-read the solution again, plug both 1 and -1 and read this reply: https://gmatclub.com/forum/m28-184547.html#p1593296
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Joined: 10 Mar 2018
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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.
Math Expert
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09 Jul 2018, 20:28
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 &nbs [#permalink] 09 Jul 2018, 20:28
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# M28-48

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