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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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12 Dec 2011, 12:31

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

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

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20 Jun 2013, 14:07

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

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

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). 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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12 Sep 2015, 07:28

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 ?

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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04 Apr 2016, 13:00

1

This post received KUDOS

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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i.e. n^2 = 1, 2, 3, 4, 5.... i.e. \(n^2 = 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}\).... NOT SIFFICIENT

Statement 2: \(\sqrt{n}\) = 1, 2, 3, 4, 5....

i.e. n = 1, 4, 9, 16, 25... SUFFICIENT

Answer: Option B
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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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24 Jan 2017, 13:55

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.

Actually why you dont ROOT both sides in (1)?

Is this because we dont know if integer is negative or not?

I mean n^2 = int and by ROOTING both sides you get n = sqrt(int) which then implies that since n can be any value is not sufficient (?). I find all this smart number approach unnecessary for this question.

Last edited by Ndkms on 25 Jan 2017, 06:34, edited 1 time in total.

(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.

Actually why you dont ROOT both sides in (1)?

Is this because we dont know if integer is negative or not?

I mean n^2 = int and by squaring both sides you get n = sqrt(int) which then implies that since n can be any value is not sufficient (?). I find all this smart number approach unnecessary for this question.

Your explanation is fine if you can easily understand n = sqrt(int) and sqrt(int) may or may NOT be an Integer therefore 1st statement is NOT SUFFICIENT

However it's not equally easy for everyone to understand as the natural bias of our mind makes us think that "if n is an integer then n^2 will also be an Integer therefore if n^2 is an Integer then n also must be an Integer"which would lead to incorrect answers.

Being a trainer I keep asking this question to the students and around 70% students get this question wrong who just begin to learn DS

Also, Smart Number approach is definitely a very good approach however the Smart number approach in isolation isn't as good enough in isolation as the Algebraic approach in isolation isn't good enough

I suggest that use of both Smart Number and Algebraic approach is the best way to perform best in any aptitude test.
_________________

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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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25 Jan 2017, 06:43

GMATinsight wrote:

Your explanation is fine if you can easily understand n = sqrt(int) and sqrt(int) may or may NOT be an Integer therefore 1st statement is NOT SUFFICIENT

However it's not equally easy for everyone to understand as the natural bias of our mind makes us think that "if n is an integer then n^2 will also be an Integer therefore if n^2 is an Integer then n also must be an Integer"which would lead to incorrect answers.

Being a trainer I keep asking this question to the students and around 70% students get this question wrong who just begin to learn DS

Also, Smart Number approach is definitely a very good approach however the Smart number approach in isolation isn't as good enough in isolation as the Algebraic approach in isolation isn't good enough

I suggest that use of both Smart Number and Algebraic approach is the best way to perform best in any aptitude test.

Imho smart number approach for this question is risky and borderline waste of time.