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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 56% (01:20) correct 44% (01:31) wrong based on 86 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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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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Answer is A.
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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1
If k is a positive integer, is k√k an integer?

(1) 1<√k<4

(2) k has exactly three factors.
The answer is B
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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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.

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

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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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Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq 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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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! Retired Moderator D
Joined: 25 Feb 2013
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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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merajul wrote:
Answer is A.
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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niks18 wrote:
merajul wrote:
Answer is A.
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 ?
Math Expert V
Joined: 02 Aug 2009
Posts: 7686
Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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merajul wrote:
niks18 wrote:
merajul wrote:
Answer is A.
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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Re: If k is a positive integer, is k^(1/2) an integer?  [#permalink]

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merajul wrote:
niks18 wrote:
merajul wrote:
Answer is A.
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.

Sent from my iPhone using GMAT Club Forum mobile app Re: If k is a positive integer, is k^(1/2) an integer?   [#permalink] 26 Dec 2017, 19:54
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