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If x is a positive integer, is root(x) an integer?

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If x is a positive integer, is root(x) an integer? [#permalink] New post 20 Feb 2010, 13:49
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If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{4x} is an integer.
(2) \sqrt{3x} is not an integer.

This is the question from GMAT Quant Review. My logic to solve this question:

[Reveal] Spoiler:
\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E.
Please advice.
[Reveal] Spoiler: OA
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Re: Is OG Quant question answer wrong? [#permalink] New post 20 Feb 2010, 14:14
alexBLR wrote:
This is the question from GMAT Quant Review:

If x is a positive integer , is \sqrt{x} an integer?

1) \sqrt{4x} is an integer.
2) \sqrt{3x} is not an integer.

My logic to solve this question:

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E.
Please advice.


IMO D...

Ques: if x is a positive integer, is \sqrt{x} an integer?

S1: \sqrt{4x} is an integer

--> 2* \sqrt{x} is an integer --> \sqrt{x} has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF

S2: \sqrt{3x} is an integer
--> \sqrt{3}*\sqrt{x} --> \sqrt{x} is not an integer as same could be a of a form of a\sqrt{3} where 'a' is a positive integer. Therefore SUFF
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Re: Is OG Quant question answer wrong? [#permalink] New post 20 Feb 2010, 14:57
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jeeteshsingh wrote:
alexBLR wrote:
This is the question from GMAT Quant Review:

If x is a positive integer , is \sqrt{x} an integer?

1) \sqrt{4x} is an integer.
2) \sqrt{3x} is not an integer.

My logic to solve this question:

\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E.
Please advice.


IMO D...

Ques: if x is a positive integer, is \sqrt{x} an integer?

S1: \sqrt{4x} is an integer

--> 2* \sqrt{x} is an integer --> \sqrt{x} has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF

S2: \sqrt{3x} is an integer
--> \sqrt{3}*\sqrt{x} --> \sqrt{x} is not an integer as same could be a of a form of a\sqrt{3} where 'a' is a positive integer. Therefore SUFF


If x is a positive integer, is \sqrt{x} an integer?

As given that x is a positive integer then \sqrt{x} is either an integer itself or an irrational number.

(1) \sqrt{4x} is an integer --> 2\sqrt{x}=integer --> 2\sqrt{x} to be an integer \sqrt{x} must be an integer or integer/2, but as x is an integer, then \sqrt{x} can not be integer/2, hence \sqrt{x} is an integer. Sufficient.

(2) \sqrt{3x} is not an integer --> if x=9, condition \sqrt{3x}=\sqrt{27} is not an integer satisfied and \sqrt{x}=3 IS an integer, BUT if x=2, condition \sqrt{3x}=\sqrt{6} is not an integer satisfied and \sqrt{x}=\sqrt{2} IS NOT an integer. Two different answers. Not sufficient.

Answer: A.

jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements can not give you TWO DIFFERENT answers to the question (as you've got).

Hope it helps.
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Re: Is OG Quant question answer wrong? [#permalink] New post 20 Feb 2010, 15:13
Bunuel wrote:
jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements can not give you TWO DIFFERENT answers to the question (as you've got).

Hope it helps.


:wall My Bad.... overlooked it... Infact today I was telling this to someone over the forum that both the statements in DS would always be in sync..

Thanks Bunuel... for pointing this out.
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Re: Is OG Quant question answer wrong? [#permalink] New post 20 Feb 2010, 15:42
When I assumed the case \sqrt{x}=2.5 I did not take into the account that x will not be an integer in this case(x=6.25). Thanks Bunuel
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Re: Integers [#permalink] New post 28 Feb 2011, 11:56
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Merging similar topics.

Similar questions:
if-x-is-a-positive-integer-is-sqrt-x-an-integer-88994.html
value-of-x-107195.html
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odd-vs-even-trick-question-106562.html
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quant-review-2nd-edition-ds-104421.html
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airthmetic-ds-108287.html

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Re: If x is a positive integer, is root(x) an integer? [#permalink] New post 05 Jun 2013, 02:55
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on roots problems: math-number-theory-88376.html

All DS roots problems to practice: search.php?search_id=tag&tag_id=49
All PS roots problems to practice: search.php?search_id=tag&tag_id=113

Tough and tricky exponents and roots questions (DS): tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky exponents and roots questions (PS): new-tough-and-tricky-exponents-and-roots-questions-125956.html

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Re: If x is a positive integer, is root(x) an integer? [#permalink] New post 05 Jun 2013, 06:58
[quote="alexBLR"]If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{4x} is an integer.
(2) \sqrt{3x} is not an integer.


(1) since \sqrt{4x} is an integer, it is the same as 2*\sqrt{x} that means \sqrt{x} also must be ans integer - sufficient

(2) \sqrt{3x} is not an integer, in this case \sqrt{x} could be or couldn't be an integer.
Case 1) x=4, where \sqrt{x} is an integer, then \sqrt{3*4}=2\sqrt{3} still not an integer
Case 2) x=5, where \sqrt{x} is NOT an integer, then \sqrt{3*5}=\sqrt{15} still not an integer.
This statement is not sufficient. Answer is A
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Re: If x is a positive integer, is root(x) an integer? [#permalink] New post 27 Dec 2013, 05:37
alexBLR wrote:
If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{4x} is an integer.
(2) \sqrt{3x} is not an integer.

This is the question from GMAT Quant Review. My logic to solve this question:

[Reveal] Spoiler:
\sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient.

\sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient.

S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement.

So I choose E as an answer.

Is there a flaw in my reasoning? OG Quant review answer to this question is different from E.
Please advice.


From First statement we get 2 sqrt (x) is an integer
Therefore sqrt (x) must be an integer

Just to elaborate a bit further:

The thing is that sqrt (x) can't be an integer/2, cause the square root of any positive integer (as stated in the givens) is always >=1. Therefore, sqrt (x) will have to be an integer.

Suff

From second setatement we get the sqrt (3x) is not an integer
Therefore, 3x is not a perfect square
or x is not perfect square that is multiple of 3
But it could very well be a perfect square
Matter is we still don't know

Just to clarify, sqrt (x) could or could not be an integer. If sqrt root (x) is an integer then sure 1.7 (integer) not an integer. But, sqrt (x) could also be say x=2, therefore sqrt (2)=1.4. Then 1.4*1.7 is clearly not an integer either. Answer A stands

Insuff

Hence A is our best choice here

Hope it helps
Let the kudos rain begin

Cheers!
J :)
Re: If x is a positive integer, is root(x) an integer?   [#permalink] 27 Dec 2013, 05:37
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