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# If k is a positive integer, is k^(1/2) an integer?

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Math Expert
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If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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24 Dec 2017, 01:24
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55% (hard)

Question Stats:

56% (00:51) correct 44% (01:07) wrong based on 81 sessions

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If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) $$1 < \sqrt{k} < 4$$

(2) k has exactly three factors.

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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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24 Dec 2017, 01:43
Imo B
St1:
1<k<16

K=4 YES
K=5 NO

Not sufficient

St2: k=4
K= 25
K = 49

Sufficient

Imo B

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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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24 Dec 2017, 07:14
Statement 1: 1 < √ k < 4
That means √ k is either 2 or 3 (k = +ve integer) , which is integer. SUFFICIENT
Statement 2: k has 3 factors. We do not know anything about it.
Lets says k = 4 = 2 X 2 X 1 ( 3 factors ) , √ k = 2 ,integer
6 = 1 X 2 X 3 ( 3 factors ) , √ k = √ 6, non-integer. INSUFFICIENT
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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24 Dec 2017, 09:51
1
If k is a positive integer, is k√k an integer?

(1) 1<√k<4

(2) k has exactly three factors.
Statement is insufficient as √k can have any value fro example √k can be 1.5 or √2
Statement is sufficient as any number which has 3 factors is a perfect square meaning its square root will be an integer
for example √36=6 , √9=3
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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24 Dec 2017, 10:41
Bunuel wrote:
If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) $$1 < \sqrt{k} < 4$$

(2) k has exactly three factors.

For $$\sqrt{k}$$ to be an integer, $$k$$ must be a perfect square.

(1) $$1 < √k < 4$$. If $$k=15$$, then $$\sqrt{k}=non-integer$$; but if $$k=36$$ then $$\sqrt{k}=integer=6$$; more than one value, insufficient.

(2) k has exactly three factors. Then $$k$$ must be a perfect square, and since $$k=any:{2^2,3^2,4^2,5^2,13^2,etc...$$, it follows that $$\sqrt{k}=integer$$, sufficient.

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If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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Updated on: 26 Dec 2017, 09:15
If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) $$1 < \sqrt{k} < 4$$

As k is positive we can safely square all terms of the inequality
squaring all:
1<k<16 => k can be any positive integer in this range and accordingly sq. root of k may or may not be an integer. Insufficient

(2) k has exactly three factors.

Only square of prime numbers will have EXACTLY three factors
E.g: 9 ==> 1,3,9
25==> 1,5,25
But 16==> 1,2,4,8,16 (more than three factors)

So, $$k= p^2$$ (p is a prime number)
$$\sqrt{k}$$= p.....hence an integer.
Sufficient

--B--
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Originally posted by TaN1213 on 26 Dec 2017, 04:59.
Last edited by TaN1213 on 26 Dec 2017, 09:15, edited 1 time in total.
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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26 Dec 2017, 09:08
Thank you Bunuel for a great question as always

I have recently given my GMAT (Q50) and through these posts want to share some tips, hacks with aspirants

Tip: Try to (algebraically) link back the statement expression to the original question variable, here we have \sqrt{k} in statement (1) and the question talks about k. So a natural way of linking is by squaring the statement expression. This may seem simple enough in this example but you would be surprised on the clearer perspective this can shed in multiple situation.

Coming to the question

Statement(1)

1 < \sqrt{k} < 4 ------> squaring the expression ---------> 1 < k < 16 (if k is 13 or 15, we get \sqrt{k} is not an integer and is k is 9, \sqrt{k} is an integer, hence ambigious information source)

Statement(2)

As others have noted, only prime squares have three unique factors (the prime square, the prime and 1). But how did we figure out, it has to be a prime square? What is the mental model to follow in the real test to not throw darts in the dark randomly...
Do a quick 'stress' test. We know that [even numbers, odd numbers, prime numbers, the number 1] are sets of positive number which generally display different characteristics. Take a example from each and see square gets us three unique factors. We are essentially doing a boundary test - taking a representative from different type of numbers and figuring out which is the ideal candidate for the question.

Hence, we go with option (B)

I have complied all my learnings, tips and hacks in a dedicated resource - mastergmatds [dot] com. Please do check it out!
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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26 Dec 2017, 09:40
merajul wrote:
Statement 1: 1 < √ k < 4
That means √ k is either 2 or 3 (k = +ve integer) , which is integer. SUFFICIENT
Statement 2: k has 3 factors. We do not know anything about it.
Lets says k = 4 = 2 X 2 X 1 ( 3 factors ) , √ k = 2 ,integer
6 = 1 X 2 X 3 ( 3 factors ) , √ k = √ 6, non-integer. INSUFFICIENT

Hi merajul

its mentioned that $$k$$ is an integer but $$\sqrt{k}$$ can be a non integer. Suppose $$k=3$$ then $$\sqrt{k}=\sqrt{3}=1.732$$. this value is between 1 & 4.

Statement 2 clearly says that $$k$$ has 3 factors. Perfect squares have 3 factors. Hence $$k$$ is a perfect square so $$\sqrt{k}$$ will be any integer
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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26 Dec 2017, 18:27
niks18 wrote:
merajul wrote:
Statement 1: 1 < √ k < 4
That means √ k is either 2 or 3 (k = +ve integer) , which is integer. SUFFICIENT
Statement 2: k has 3 factors. We do not know anything about it.
Lets says k = 4 = 2 X 2 X 1 ( 3 factors ) , √ k = 2 ,integer
6 = 1 X 2 X 3 ( 3 factors ) , √ k = √ 6, non-integer. INSUFFICIENT

Hi merajul

its mentioned that $$k$$ is an integer but $$\sqrt{k}$$ can be a non integer. Suppose $$k=3$$ then $$\sqrt{k}=\sqrt{3}=1.732$$. this value is between 1 & 4.

Statement 2 clearly says that $$k$$ has 3 factors. Perfect squares have 3 factors. Hence $$k$$ is a perfect square so $$\sqrt{k}$$ will be any integer

Hi niks18

thnx. I missed that part in A.

However, for B , I am not convinced. for example

4 = 2^2 X 1^0 , No of factors = (2+1)(0+1) = 3
9 = 3^2 X 1^0 , No of factors = (2+1)(0+1) = 3
This applies for 16,25,49...
But ,
36 = 2^2 X 3^2 X 1^0 , factors = (2+1)(2+1)(0+1)=9
64 = 2^6 X 1^0, no of factors = ( 6+1)(0+1)=7
81 , factors = 5
100,factors = (2+1)(2+1)(0+1) = 9
.
.
.
and so on.

How do you explain this ?
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Joined: 02 Aug 2009
Posts: 6519
Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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26 Dec 2017, 19:26
merajul wrote:
niks18 wrote:
merajul wrote:
Statement 1: 1 < √ k < 4
That means √ k is either 2 or 3 (k = +ve integer) , which is integer. SUFFICIENT
Statement 2: k has 3 factors. We do not know anything about it.
Lets says k = 4 = 2 X 2 X 1 ( 3 factors ) , √ k = 2 ,integer
6 = 1 X 2 X 3 ( 3 factors ) , √ k = √ 6, non-integer. INSUFFICIENT

Hi merajul

its mentioned that $$k$$ is an integer but $$\sqrt{k}$$ can be a non integer. Suppose $$k=3$$ then $$\sqrt{k}=\sqrt{3}=1.732$$. this value is between 1 & 4.

Statement 2 clearly says that $$k$$ has 3 factors. Perfect squares have 3 factors. Hence $$k$$ is a perfect square so $$\sqrt{k}$$ will be any integer

Hi niks18

thnx. I missed that part in A.

However, for B , I am not convinced. for example

4 = 2^2 X 1^0 , No of factors = (2+1)(0+1) = 3
9 = 3^2 X 1^0 , No of factors = (2+1)(0+1) = 3
This applies for 16,25,49...
But ,
36 = 2^2 X 3^2 X 1^0 , factors = (2+1)(2+1)(0+1)=9
64 = 2^6 X 1^0, no of factors = ( 6+1)(0+1)=7
81 , factors = 5
100,factors = (2+1)(2+1)(0+1) = 9
.
.
.
and so on.

How do you explain this ?

Perfect square of PRIME numbers have 3 factors..
So 3^2,5^2,7^2....
B talks of this set of numbers
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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26 Dec 2017, 19:54
merajul wrote:
niks18 wrote:
merajul wrote:
Statement 1: 1 < √ k < 4
That means √ k is either 2 or 3 (k = +ve integer) , which is integer. SUFFICIENT
Statement 2: k has 3 factors. We do not know anything about it.
Lets says k = 4 = 2 X 2 X 1 ( 3 factors ) , √ k = 2 ,integer
6 = 1 X 2 X 3 ( 3 factors ) , √ k = √ 6, non-integer. INSUFFICIENT

Hi merajul

its mentioned that $$k$$ is an integer but $$\sqrt{k}$$ can be a non integer. Suppose $$k=3$$ then $$\sqrt{k}=\sqrt{3}=1.732$$. this value is between 1 & 4.

Statement 2 clearly says that $$k$$ has 3 factors. Perfect squares have 3 factors. Hence $$k$$ is a perfect square so $$\sqrt{k}$$ will be any integer

Hi niks18

thnx. I missed that part in A.

However, for B , I am not convinced. for example

4 = 2^2 X 1^0 , No of factors = (2+1)(0+1) = 3
9 = 3^2 X 1^0 , No of factors = (2+1)(0+1) = 3
This applies for 16,25,49...
But ,
36 = 2^2 X 3^2 X 1^0 , factors = (2+1)(2+1)(0+1)=9
64 = 2^6 X 1^0, no of factors = ( 6+1)(0+1)=7
81 , factors = 5
100,factors = (2+1)(2+1)(0+1) = 9
.
.
.
and so on.

How do you explain this ?

Hi @merajul,

Statement B mentions that k has EXACTLY 3 factors and as explained by @chetan2u, only perfect squares of PRIME NUMBERS have exactly 3 factors because a prime is divisible only by 1 & itself only.

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Re: If k is a positive integer, is k^(1/2) an integer? &nbs [#permalink] 26 Dec 2017, 19:54
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