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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:


(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
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Re: M04-07 [#permalink] New post  02 Dec 2014, 23:09
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I got this question wrong. According to rules for the GMAT, if x is under a squareroot, it's possible value cannot a negative. so, because of this rule, my thinking was that x could only equal 1. Am i missing something here ?
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Bunuel
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Re: M04-07 [#permalink] New post  03 Dec 2014, 04:26
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shresthas wrote:
I got this question wrong. According to rules for the GMAT, if x is under a squareroot, it's possible value cannot a negative. so, because of this rule, my thinking was that x could only equal 1. Am i missing something here ?


Yes, on the GMAT, the expression under the even roots, for example under the square root, must be non-negative. (2) says that \(2\sqrt{x^2}=prime\). Notice that the expression under the square root is the square of a number (x^2), and the square of a number cannot be negative. But x itself CAN be negative, for example: if x = -1, then \(2\sqrt{(-1)^2}=2\sqrt{1}=2=prime\).

Hope it's clear.
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Re: M04-07 [#permalink] New post  28 May 2015, 11:26
Bunuel wrote:
Official Solution:

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


Hi!
Is it possible for a prime to be negative?
|23x| is either positive or negative. Provided that |23x| is prime x has to be equal 1 so that 23x>0. What do I miss?

Thanks.
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Bunuel
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Re: M04-07 [#permalink] New post  28 May 2015, 11:28
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dbereza wrote:
Bunuel wrote:
Official Solution:

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


Hi!
Is it possible for a prime to be negative?
|23x| is either positive or negative. Provided that |23x| is prime x has to be equal 1 so that 23x>0. What do I miss?

Thanks.


Absolute value of a number cannot be negative.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

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Re: M04-07 [#permalink] New post  30 Jun 2015, 10:23
Got it wrong as I missed that x can be negative too and prime number can only be positive integer
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Re: M04-07 [#permalink] New post  20 Sep 2015, 09:57
Bunuel wrote:
Official Solution:

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

Hi Bunuel,
In gmat, if \({x^2}\) = 25, then the value of x will be +5 not -5.
Here, why are we considering -ve value of x? (from statement 2!)
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Bunuel
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Re: M04-07 [#permalink] New post  20 Sep 2015, 20:34
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scofield1521 wrote:
Bunuel wrote:
Official Solution:

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

Hi Bunuel,
In gmat, if \({x^2}\) = 25, then the value of x will be +5 not -5.
Here, why are we considering -ve value of x? (from statement 2!)


That's not true.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive 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.

Hope it's clear.
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Re: M04-07 [#permalink] New post  08 Apr 2020, 07:40
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.


x=?

\(|23x|\) is a prime number.

Let's think about this. x cannot be 0. x cannot be any value greater than or equal to 2 because that will result in 23x not being Prime.

So the only value x can take is 1. But then be careful. \(|23x|\) is prime. So x can be 1 or -1.

Not Sufficient.

(2) \(2\sqrt{x^2}\) is a prime number.

Because we did efforts in Statement 1, we can clearly see that both values of x work here too.

Not Sufficient.

Combined:

Still x can be 1 or -1. No new information to eliminate any of the two values.

Answer E.
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Re: M04-07 [#permalink] New post  08 Apr 2020, 09:34
In statement 2 I wanted the square root to cancel the square of x so I wanted to be left with 2*1

Why isn’t this correct?

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Re: M04-07 [#permalink] New post  08 Apr 2020, 09:38
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6mboyjam wrote:
In statement 2 I wanted the square root to cancel the square of x so I wanted to be left with 2*1

Why isn’t this correct?

Posted from my mobile device


I guess you wanted to write Recall that \(\sqrt{x^2}=x\).

That's not correct because \(\sqrt{x^2}=|x|\), not x.

Why?

The point here is that square root function cannot give negative result: which 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: M04-07 [#permalink] New post  29 Dec 2020, 13:55
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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) \(|23x|\) is within absolute value of x can be -1, or 1; Insufficient.

(2) x can be -1 or 1, for either case the \(2\sqrt{x^2}\) will be prime.

For example
\(x=-1; 2\sqrt{(-1)^2}; 2\sqrt{1}=2\)

\(x=1; 2\sqrt{1^2}; 2\sqrt{1}=2\); Insufficient.

Considering both:
Still, x can be -1 or 1; Insufficient.

The answer is \(E\)
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Re: M04-07 [#permalink] New post  29 Dec 2020, 14:59
A small clarification:

|23x| means |23* x| but not 23X ( as in 231,233,239)?
if we use variable with number then by default it means multiplication? Am i right? Bunuel
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Re: M04-07 [#permalink] New post  29 Dec 2020, 15:09
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imSKR wrote:
A small clarification:

|23x| means |23* x| but not 23X ( as in 231,233,239)?
if we use variable with number then by default it means multiplication? Am i right? Bunuel


Right. If 23x were a three-digit number it would have been mentioned explicitly. Without that, 23x can only be 23*x, since only multiplication sign (*) is usually omitted.
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Re: M04-07 [#permalink] New post  23 Feb 2023, 06:32
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Duplicate of M28-48. Unpublished....................
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