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

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enigma123
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If k is a positive integer, is square root of k an integer? [#permalink] New post   07 Mar 2012, 15:55
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If k is a positive integer, is \(\sqrt{k}\) an integer?

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

(2) \(k^2 < 16\)

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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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C
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Bunuel
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Re: Integer K [#permalink] New post   07 Mar 2012, 16:06
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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.

Answer: C.
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Re: If k is a positive integer, is square root of k an integer? [#permalink] New post   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] New post   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.

Answer: C.


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...
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Re: If k is a positive integer, is square root of k an integer? [#permalink] New post   31 Oct 2020, 08:12
If k is a positive integer, is \(\sqrt{k}\) an integer?

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

\(1 < k < 9\)

K could or could not be an integer. INSUFFICIENT.

(2) \(k^2 < 16\)

Since k must be a positive integer, k can be 1, 2, or 3.

If k is 1, then \(\sqrt{k}\) is an integer. If k = 2 or 3, \(\sqrt{k}\) is not an integer. INSUFFICIENT.

(1&2) \(1 < k < 9\)
k must be 2 or 3. For either number, \(\sqrt{k}\) is NOT an integer. SUFFICIENT.

Answer is C.
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Re: If k is a positive integer, is square root of k an integer? [#permalink]
31 Oct 2020, 08:12
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Bunuel
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