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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, 10:27
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87% (01:45) correct 13% (01:01) wrong based on 148 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, 11:08
2. The sqr(d) is a square of an integer, thus we do know that a square of a ninteger must be an integer- SQR(D) is an integer- sufficient.
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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 11:19
I dont know why i cant see this, Would you be able to help visualize with numbers..?
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Re: Quant Review 2nd Edition: DS 115 [#permalink]

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07 Nov 2010, 12:39
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, 12:44
2
KUDOS
Expert's post
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, 13:49
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, 16:07
Just put two values for positive integer variable D. For ex- D=15 or D=16.
Re: Quant Review 2nd Edition: DS 115   [#permalink] 07 Nov 2010, 16:07
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# If d is a positive integer is d^1/2 an intger

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