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# If k is a positive integer, is square root of k an integer?

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If k is a positive integer, is square root of k an integer?  [#permalink]

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07 Mar 2012, 15:55
2
1
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Difficulty:

75% (hard)

Question Stats:

44% (01:23) correct 56% (01:07) wrong based on 184 sessions

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

(1) $$1 < \sqrt{k} < 3$$

(2) $$k^2 < 16$$

How come the answer is C?

Statement 1 - The only value K can take is 2. So this is sufficient.

Statement 2 - is not sufficient as -4 < k < 4.

Where I am incorrect guys?

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07 Mar 2012, 16:06
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2
enigma123 wrote:
If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) $$1 < \sqrt{k} < 3$$

(2) $$k^2 < 16$$

How come the answer is C?

Statement 1 - The only value K can take is 2. So this is sufficient.

Statement 2 - is not sufficient as -4 < k < 4.

Where I am incorrect guys?

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

(1) $$1 < \sqrt{k} < 3$$. Square the given inequality: $$1<k<9$$ --> $$k$$ can be 2, 3, 4, 5, 6, 7, or 8. If $$k=4$$ then $$\sqrt{k}$$ is an integer, for other values it's not an integer. Not sufficient.

(2) $$k^2 < 16$$ --> $$-4<k<4$$, since it's also given that $$k$$ is a positive integer, then $$k$$ can be 1, 2 or 3. If $$k=1$$ then the answer is YES but if $$k$$ is 2 or 3 then the answer is NO. Not sufficient.

(1)+(2) Intersection of the values of $$k$$ from (1) and (2) is 2 and 3, $$\sqrt{k}$$ is NOT an integer for either of them. Sufficient.

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

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13 Mar 2016, 08:50
The other way to look at this question is to find the values of k =>
statement 1 => k=> 2,3,4,5,6,7,8
statement 2 => k=> 1,2,3
combining them k=> 2,3
so √k will never be an integer
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Re: If k is a positive integer, is square root of k an integer?  [#permalink]

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06 Apr 2017, 12:42
Bunuel wrote:
enigma123 wrote:
If k is a positive integer, is $$\sqrt{k}$$ an integer?

(1) $$1 < \sqrt{k} < 3$$

(2) $$k^2 < 16$$

How come the answer is C?

Statement 1 - The only value K can take is 2. So this is sufficient.

Statement 2 - is not sufficient as -4 < k < 4.

Where I am incorrect guys?

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

(1) $$1 < \sqrt{k} < 3$$. Square the given inequality: $$1<k<9$$ --> $$k$$ can be 2, 3, 4, 5, 6, 7, or 8. If $$k=4$$ then $$\sqrt{k}$$ is an integer, for other values it's not an integer. Not sufficient.

(2) $$k^2 < 16$$ --> $$-4<k<4$$, since it's also given that $$k$$ is a positive integer, then $$k$$ can be 1, 2 or 3. If $$k=1$$ then the answer is YES but if $$k$$ is 2 or 3 then the answer is NO. Not sufficient.

(1)+(2) Intersection of the values of $$k$$ from (1) and (2) is 2 and 3, $$\sqrt{k}$$ is NOT an integer for either of them. Sufficient.

what a brilliant explanation...I answered E, but now I see why I made the mistake...
i thought that k^2 = 4, therefore sqrt(k) can be either a non-integer or an integer...
Re: If k is a positive integer, is square root of k an integer? &nbs [#permalink] 06 Apr 2017, 12:42
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# If k is a positive integer, is square root of k an integer?

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