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

Question

Is n an integer?

(1)  n^2  is an integer

(2)  \sqrt{n}  is an integer

Bunuel · Answer

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.

Answer: B.

vwjetty · Answer

oh wow. i feel dumb now. for some reason i was thinking something else. tricky tricky. thx. +1 kudo's good sir.

b2bt · Answer

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

Bunuel · Answer

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?

b2bt · Answer

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

Bunuel · Answer

The square root of 3 is a number which when squared gives 3.

kunal555 · Answer

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 ?

Bunuel · Answer

The square root of 2 is a number (whatever it is) which when squared gives 2.

GMATinsight · Answer

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

Answer: Option B

Ndkms · Answer

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.

GMATinsight · Answer

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.

alice7 · Answer

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

Bunuel · Answer

Why cannot n^2 be 3? In this case n=\sqrt{3} or -\sqrt{3} , so in this case n is an irrational number. For more on number theory (for example on irrational numbers) check the following link: https://gmatclub.com/forum/math-number- ... 88376.html Hope it helps.

JeffTargetTestPrep · Answer

We need to determine whether n is an integer.

Statement One Alone:

n^2 is an integer.

If n^2 is an integer, n may or may not be an integer. For instance, if n^2 = 4, then n is an integer (since n = 2 or -2). However, if n^2 = 5, then n is not an integer (since n = √5 or -√5). Statement one is not sufficient to answer the question.

Statement Two Alone:

√n is an integer.

In order for √n to be an integer, n must be an integer. This is because n = (√n)^2, and any integer squared is also an integer. Statement two is sufficient to answer the question.

Answer: B

GMATinsight · Answer

A proper way to answer any DS question

Check the video solution of this question

https://www.youtube.com/watch?v=UEoM8XPx1S0

BLTN · Answer

Interesting but cunning question

Statement 1 ( \sqrt{2} )^2 = 2 yet, \sqrt{2} is not an integer Conversely, 2^2 =4 and 2 is an integer. Statement 2 is sufficient by its own.

rak08 · Answer

can solve like this :

n^2 = integer
take root both sides
n = root of integer
which may or may not be an integer.

however 
if sq. root n = integer
squaring both sides.
n = integer ^2
n = integer.
hence B

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

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