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Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an

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Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer
[Reveal] Spoiler: OA

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Last edited by Bunuel on 05 Apr 2012, 03:48, edited 1 time in total.
Edited the question and added the OA
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vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.


Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.
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New post 31 Aug 2010, 14:20
oh wow. i feel dumb now. for some reason i was thinking something else. tricky tricky. thx. +1 kudo's good sir.
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New post 12 Dec 2011, 13:31
Answer is indeed B.

stmt 1: says that n^2 is an integer. So n could be 4, which n^16 or n could be 5.477 and n^2 would be 30. This gives us two different options so insuff

stmt 2: says that sqrt n is an integer. that means n is the square of an (or integer x integer) which will always give you an integer. suff
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New post 23 Dec 2011, 01:16
Is n an integer?

(1) n^2 is an integer

(2) sqrt(n) is an integer

S1: If n^2 is an integer, is n an integer?

Is n rational or irrational?

If rational, then n can be expressed as p/q,
where p and q are non-zero integers.
Further, if n^2 = (p/q)^2 is an integer, n = p/q is an integer.

If irrational, then n could be the non-integer square root of an integer, whose square is an integer.

S2: An integer multiplied by another integer is an integer.
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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New post 05 Apr 2012, 04:31
Thank you for the explanations. I know have an understanding of the question.
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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New post 20 Jun 2013, 15:07
Stament 1 : n^2=integer
n=+/- sqrt(integer)
=> Sqrt of integer may or may not be integer depending on whether chose integer is perfect square
or not. insufficient

Statement 2: sqrt (n) = integer
=> Squaring both sides
=> n= integer^2 = some other integer (as intg*intg=intg)
= sufficient

Ans B
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Re: N [#permalink]

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New post 30 May 2014, 02:27
Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.


Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.


If we take \(\sqrt{3}\) = 1.732050807568877
and square it, we still get 2.999999999999
which is not an integer. That's the reason I marked D :( :roll:

Last edited by b2bt on 30 May 2014, 04:59, edited 1 time in total.
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Re: N [#permalink]

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New post 30 May 2014, 03:21
Expert's post
b2bt wrote:
Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.


Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.


If we take \sqrt{3} = 1.732050807568877
and square it, we still get 2.999999999999
which is not an integer. That's the reason I marked D :( :roll:


Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Please read.

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?
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Re: N [#permalink]

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New post 30 May 2014, 04:54
Quote:

Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Please read.

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?


How can square of non-integer be an integer?
1.73 is a non- integer and so is its square
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New post 30 May 2014, 05:06
Expert's post
b2bt wrote:
Quote:

Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Please read.

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?


How can square of non-integer be an integer?
1.73 is a non- integer and so is its square


The square root of 3 is a number which when squared gives 3.
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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New post 12 Sep 2015, 08:28
Statement 1 says \(n^2\) is an integer.
According to the solution, \((\sqrt{2})^2\) is an integer.
But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx.
And when we square it we would get 1.9999999 approx.
We do not get an integer on squaring an irrational number but rather a value close to an integer.
So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\)
For this reason, i marked D.
Can someone please explain why is my thinking wrong and why are we taking approximate values ?
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New post 13 Sep 2015, 03:14
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kunal555 wrote:
Statement 1 says \(n^2\) is an integer.
According to the solution, \((\sqrt{2})^2\) is an integer.
But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx.
And when we square it we would get 1.9999999 approx.
We do not get an integer on squaring an irrational number but rather a value close to an integer.
So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\)
For this reason, i marked D.
Can someone please explain why is my thinking wrong and why are we taking approximate values ?


The square root of 2 is a number (whatever it is) which when squared gives 2.
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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New post 04 Apr 2016, 14:00
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Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.


Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO. <<<< n will be square root of the value. So it can be an integer or non-integer, depending on the value whether it is a perfect square or not. So, not sufficient.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.




I have added few more explanation for the 1st option, as i got this question wrong because of that problem. Hope you guys will not repeat the same mistake. #kudos
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an   [#permalink] 04 Apr 2016, 14:00
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