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# Is k the square of an integer?

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Joined: 27 Feb 2012
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Is k the square of an integer?  [#permalink]

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Updated on: 05 Sep 2017, 01:08
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Difficulty:

15% (low)

Question Stats:

78% (00:54) correct 22% (01:04) wrong based on 143 sessions

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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

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.
Edited the question.
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Re: Is k the square of an integer?  [#permalink]

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18 Dec 2012, 23:18
1
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.

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Re: Is k the square of an integer?  [#permalink]

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19 Dec 2012, 03:05
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.

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Re: Is k the square of an integer?  [#permalink]

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25 Jul 2018, 14:09
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Re: Is k the square of an integer? &nbs [#permalink] 25 Jul 2018, 14:09
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