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Is k the square of an integer?

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jsphcal
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Is k the square of an integer? [#permalink] New post  Updated on: 05 Sep 2017, 01:08
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Is k the square of an integer?

(1) \(k= t^2 * q^6 *r^{10}\), where t, q and r are integers

(2) \(k= t^2 * m^{15}\), where t and m are integers
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Originally posted by jsphcal on 18 Dec 2012, 23:04.
Last edited by Bunuel on 05 Sep 2017, 01:08, edited 4 times in total.
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Re: Is k the square of an integer? [#permalink] New post  18 Dec 2012, 23:18
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jsphcal wrote:
Is k the square of an integer?

(1) \(k= t ^2 * q^6 * r^10\) , where t, q and r are integers


(2) \(k= r ^2 * m^{15}\) , where r and m are integers


Please correct the typo in statement 1... :)

1) we have\(k={(tq^3r^5)}^2\). Since t,q and r are intergers, k is the square of an integer. Sufficient.

2)m can take several values for k to not be the square of an integer. If m =1, k is the square of an integer. Insufficient.

Answer is A.
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Re: Is k the square of an integer? [#permalink] New post  19 Dec 2012, 03:05
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Is k the square of an integer?

We need to determine whether k is a perfect square.

(1) k = t^2*q^6*r^10, where t, q, and r are integers --> \(k=(tq^3r^5)^2=integer^2\). Sufficient.

(2) k = t^2*m^15, where t and m are integers --> if m is a perfect square itself (say if m=n^2 for some integer n), then k will also be a perfect square but if m is NOT a perfect square, then k also will NOT be a perfect square. For example consider m=1 and m=2. Not sufficient.

Answer: A.
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Re: Is k the square of an integer? [#permalink] New post  18 Aug 2018, 02:38
it's easy to miss that m can be zero or a squre.
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Re: Is k the square of an integer? [#permalink] New post  07 Feb 2022, 22:25
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A square of a number is derived when the number is multiplied by itself-> it has 2 copies of a number in a product

For K to be a square of an integer, it has to be an integer^2

1. k=t^2∗q^6∗r^10 where t, q and r are integers

sqrt(k) = sqrt(t^2∗q^6∗r^10) = t*q^3*r^5-> hence sqrt(K) is an integer -> sufficient

2.k=t^2*m^15 , where t and m are integers
sqrt(k)=sqrt(t^2*m^15) = t*m^7*sqrt(m)-> now in this case we cannot say whether k is a square of an integer , if m=3, then K is not the square of an integer, if m=4 then K is a square of an integer-> hence insufficient
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Re: Is k the square of an integer? [#permalink] New post  16 Dec 2022, 04:02
Solved this question using no of factors concept. No of factors of a perfect square is odd. But the reverse is not true right?
Can someone please confirm. Because I got option b wrong.
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Re: Is k the square of an integer? [#permalink] New post  27 Dec 2022, 02:32
Can someone clarify the doubt above?
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Re: Is k the square of an integer? [#permalink] New post  27 Dec 2022, 02:42
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TBT wrote:
Solved this question using no of factors concept. No of factors of a perfect square is odd. But the reverse is not true right?
Can someone please confirm. Because I got option b wrong.


The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square.


From (2) we cannot say whether the number of factors of k is even or odd because we don't know the number of factors of m. Check the solution here: https://gmatclub.com/forum/is-k-the-squ ... l#p1158051
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Re: Is k the square of an integer? [#permalink] New post  27 Dec 2022, 03:43
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bunuel
I got my mistake. I assumed m cannot be a perfect square itself. Thankyou!
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Re: Is k the square of an integer? [#permalink]
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