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# M28-48

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Math Expert
Joined: 02 Sep 2009
Posts: 58420

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16 Sep 2014, 01:44
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35% (medium)

Question Stats:

66% (01:06) correct 34% (01:15) wrong based on 144 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
1
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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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Joined: 14 Oct 2015
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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
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Joined: 23 Nov 2014
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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
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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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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
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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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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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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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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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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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08 Jan 2019, 07:40
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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09 Jan 2019, 01:14
Bunuel 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.

#1 :
x can be +/-1 so insufficeint
#2:
again x can be +/-1 in sufficeint

IMO E
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02 Oct 2019, 06:35
Hi
Could anyone please explain here that when in second statement: 2 sqrt(x2) is said to be prime number, doesn't that mean we have to take value of x = +1
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02 Oct 2019, 06:40
bhimeshsharma wrote:
Hi
Could anyone please explain here that when in second statement: 2 sqrt(x2) is said to be prime number, doesn't that mean we have to take value of x = +1

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Re: M28-48   [#permalink] 02 Oct 2019, 06:40
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# M28-48

Moderators: chetan2u, Bunuel