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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 [#permalink]

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31 Aug 2010, 13:04
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Is n an integer?

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

(2) $$\sqrt{n}$$ is an integer
[Reveal] Spoiler: OA

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Last edited by Bunuel on 29 Jun 2017, 10:40, edited 2 times in total.
Edited the question and added the OA

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

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31 Aug 2010, 13:18
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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) $$\sqrt{n}$$ is an integer --> $$\sqrt{n}=integer$$ --> $$n=integer^2=integer$$. Sufficient.

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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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31 Aug 2010, 13:20
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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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

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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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23 Dec 2011, 00:16
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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05 Apr 2012, 03:31
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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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

Ans 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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30 May 2014, 01:27
Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

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

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.

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

Last edited by b2bt on 30 May 2014, 03:59, edited 1 time in total.

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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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30 May 2014, 02:21
b2bt wrote:
Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

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

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.

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

$$\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...$$. It's an irrational number, it goes on forever. Anyway, what's your 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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30 May 2014, 03:54
Quote:

$$\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...$$. 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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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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30 May 2014, 04:06
b2bt wrote:
Quote:

$$\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...$$. 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 ?

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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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13 Sep 2015, 02:14
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Expert's post
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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04 Apr 2016, 13:00
1
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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

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.

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]

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17 Jul 2016, 20:56
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

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

Required: Is n an integer?

Statement 1: n^2 is an integer
If n = 2, n^2 = 4
If n = √2, n^2 = 2
INSUFFICIENT

Statement 2: √n is an integer.
If √n is an integer, then (√n)^2 = n
This will also be an integer.
SUFFICIENT

Correct 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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03 Aug 2016, 20:28
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

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

Statement 1: n^2 = Integer

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

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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

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.

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.

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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, 05:53
Ndkms wrote:
Bunuel wrote:
vwjetty wrote:
Is n an integer?

(1) n^2 is an integer

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

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.

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.

There are GMAT questions that require you to know how to ROOT both sides (like this one for example if-x-0-what-is-the-value-of-x-226745.html).

Obviously there are "many roads that lead to Rome" although if someone doesn't know how to solve (1) with "ROOTING" means that has holes in algebra.

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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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28 Jan 2017, 09:45
Hi, Bunel how could n2 could be 3 in your assumption. I think it has to be 1,2,4,9 which is sufficient.

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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an   [#permalink] 28 Jan 2017, 09:45

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