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# If k is a positive integer, is k an integer?

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Joined: 07 Feb 2015
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If k is a positive integer, is k an integer?  [#permalink]

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Updated on: 23 Jun 2015, 00:01
2
00:00

Difficulty:

15% (low)

Question Stats:

79% (01:03) correct 21% (00:53) wrong based on 91 sessions

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

(1) k is even.
(2) k = $$\sqrt{m}$$, where m is an integer.

Originally posted by gmatser1 on 22 Jun 2015, 16:35.
Last edited by Harley1980 on 23 Jun 2015, 00:01, edited 1 time in total.
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Re: If k is a positive integer, is k an integer?  [#permalink]

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22 Jun 2015, 16:37
gmatser1 wrote:
If k is a positive integer, is \sqrt{k} an integer?

(1) k is even.
(2) k = \sqrt{m}, where m is an integer.

I understand how (1) and (2) is not sufficient, but I don't get how they are not sufficient together. If k is even and k is the square root of m, shouldn't that mean that the square root of k is never an integer since 2 will be in k?
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Re: If k is a positive integer, is k an integer?  [#permalink]

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22 Jun 2015, 16:41
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When you take the square root of an even number, sometimes you get an integer, sometimes not - √2 is not an integer, but √4 is an integer. So S1 is not sufficient.

Every positive integer is the square root of some other integer (2 is the square root of 4, 3 is the square root of 9, etc) so Statement 2 tells us nothing whatsoever, and since S1 is not sufficient, the answer must be E.
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Re: If k is a positive integer, is k an integer?  [#permalink]

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23 Jun 2015, 00:03
2
gmatser1 wrote:
gmatser1 wrote:
If k is a positive integer, is \sqrt{k} an integer?

(1) k is even.
(2) k = \sqrt{m}, where m is an integer.

I understand how (1) and (2) is not sufficient, but I don't get how they are not sufficient together. If k is even and k is the square root of m, shouldn't that mean that the square root of k is never an integer since 2 will be in k?

Hello gmatser1
m can be equal to 16 and then k = 4 -> integer
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Re: If k is a positive integer, is k an integer?  [#permalink]

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23 Jun 2015, 19:57
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Hi gmatser1,

This question can be solved rather quickly by TESTing VALUES.

We're told that K is a POSITIVE INTEGER. We're asked if √K is an integer. This is a YES/NO question.

Fact 1: K is EVEN

IF....
K = 2
√2 is NOT an integer, so the answer to the question is NO

IF...
K = 4
√4 = 2 and that IS an integer, so the answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: K =√M where M is an INTEGER

IF....
K = 2, M = 4
√2 is NOT an integer, so the answer to the question is NO

IF...
K = 4, M = 16
√4 = 2 and that IS an integer, so the answer to the question is YES
Fact 2 is INSUFFICIENT

Combined, we don't need to do any additional work. We already have 2 possibilities that fit BOTH Facts (one gives us a NO answer and one gives us a YES answer).
Combined, INSUFFICIENT

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Re: If k is a positive integer, is k an integer?  [#permalink]

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12 Jul 2016, 20:39
m can be equal to 16 and then k = 4 -> integer
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Re: If k is a positive integer, is k an integer?  [#permalink]

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13 Jul 2016, 05:59
gmatser1 wrote:
If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) k is even.
(2) k = $$\sqrt{m}$$, where m is an integer.

(1) k is even.
k can be 2, 4, 6, 8, 10.... and hence $$\sqrt{k}$$ may or may not be an integer.

(2) k = $$\sqrt{m}$$, where m is an integer.

K can be 1, 2, 3, 4, .... and hence $$\sqrt{k}$$ may or may not be integer

combining both statements

k can be 2, 4... and hence $$\sqrt{k}$$ may or may not be an integer.

Both statements together are insufficient. E is the answer
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Re: If k is a positive integer, is k an integer?  [#permalink]

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11 Oct 2017, 14:50
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