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If d is a positive integer is d^1/2 an intger

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If d is a positive integer is d^1/2 an intger [#permalink]

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07 Nov 2010, 09:27
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83% (00:39) correct 17% (00:55) wrong based on 239 sessions

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

(1) $$d$$ is the square of an integer
(2) $$\sqrt{d}$$ is the square of an integer

[Reveal] Spoiler:
I have rephrased the question to basically state, is d a perfect square?

1) If d is the square of an integer this is saying that, an integer, when multiplied by itself = d --> Sufficient

2) I had a little trouble rephrasing this to be more understandable, can anybody clarify?
[Reveal] Spoiler: OA

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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 11:39
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Hi,

Correct me, If I am wrong,

Quote:
If d is a positive integer is \sqrt{d}an integer?

The Integer set contains, all the whole number and the negatives of all the natural number ie -1,-2,0,1,2

As per your statement, If Suppose 3 is a positive integer, \sqrt{3} will also be positive integer, which is not true as sqrt3 is approx 1.73, which by definition doesn't fall in integer set.

If & Only If d is a perfect square, the \sqrt{d} will a integer as in 121 will give +11, -11.

Do let me know, If I am missing something or mis-stated something.

Regards,
Tushar

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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 11:44
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jscott319 wrote:
If $$d$$ is a positive integer is $$\sqrt{d}$$an integer?

1) $$d$$ is the square of an integer
2) $$\sqrt{d}$$ is the square of an integer

I have rephrased the question to basically state, is d a perfect square?

1) If d is the square of an integer this is saying that, an integer, when multiplied by itself = d --> Sufficient

2) I had a little trouble rephrasing this to be more understandable, can anybody clarify?

Given: $$d=integer$$. Question: is $$\sqrt{d}=integer$$? or as you correctly rephrased is $$d$$ a perfect square: is $$d=integer^2$$?

(1) $$d$$ is the square of an integer --> $$d=integer^2$$ --> directly gives an answer. Sufficient.
(2) $$\sqrt{d}$$ is the square of an integer --> $$\sqrt{d}=integer^2=integer$$ (as square of an integer is an integer). Sufficient.

Similar questions:
i-cant-understand-how-the-oa-is-101475.html?hilit=irrational
algebra-ds-101464.html

Hope it helps.
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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 12:49
Thank you that does help. Not sure why I couldnt process it ha. Tricky wording. Thanks for the link as well +1

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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 15:07
Just put two values for positive integer variable D. For ex- D=15 or D=16.

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Re: If d is a positive integer is d^1/2 an intger [#permalink]

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24 Nov 2017, 23:30
jscott319 wrote:
If $$d$$ is a positive integer is $$\sqrt{d}$$ an integer?

(1) $$d$$ is the square of an integer
(2) $$\sqrt{d}$$ is the square of an integer

[Reveal] Spoiler:
I have rephrased the question to basically state, is d a perfect square?

1) If d is the square of an integer this is saying that, an integer, when multiplied by itself = d --> Sufficient

2) I had a little trouble rephrasing this to be more understandable, can anybody clarify?

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-d-is-a-po ... 67950.html
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Re: If d is a positive integer is d^1/2 an intger   [#permalink] 24 Nov 2017, 23:30
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