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]

20 Feb 2010, 14:49

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47% (00:38) wrong

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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]

20 Feb 2010, 15:14

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]

20 Feb 2010, 15:57

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jeeteshsingh wrote:

alexBLR wrote:

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]

20 Feb 2010, 16:13

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.

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]

20 Feb 2010, 16:42

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]

28 Feb 2011, 12:56

Merging similar topics.

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quant-review-2nd-edition-ds-104421.html
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Hope it helps.
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Re: If x is a positive integer, is root(x) an integer? [#permalink]

05 Jun 2013, 03:55

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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All DS roots problems to practice: search.php?search_id=tag&tag_id=49
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Tough and tricky exponents and roots questions (DS): tough-and-tricky-exponents-and-roots-questions-125967.html
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Re: If x is a positive integer, is root(x) an integer? [#permalink]

05 Jun 2013, 07:58

[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] 05 Jun 2013, 07:58

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

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