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If c and d are positive, is d an integer ?

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If c and d are positive, is d an integer ? [#permalink] New post 22 Jul 2012, 17:17
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If c and d are positive, is d an integer ?

(1) c=d^3
(2) d=\sqrt{c}
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Jul 2012, 00:40, edited 2 times in total.
Edited the question.
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Re: If c & d are positive, is d an integer? [#permalink] New post 22 Jul 2012, 19:26
pathakshashi wrote:
If c & d are positive, is d an integer ?
i) c=d^3
ii) d=\sqrt{c}


My answer is C and below is the explanation.
i) c=d^3 -> c & d can have multiple fraction values to satisfy this equation. Hence Insuff.
ii) d=\sqrt{c} -> Here as well, c & d can have multiple fraction values to satisfy this equation. Hence Insuff.

Putting together -> d = square_root (d^3) and this equation can have only one value, which is d =1. Hence C is the answer.

Cheers!
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Re: If c and d are positive, is d an integer ? [#permalink] New post 23 Jul 2012, 00:50
Expert's post
If c and d are positive, is d an integer ?

(1) c=d^3. If c is a perfect cube (for example 1^3=1, 2^3=8, 3^3=27, ...), then the answer is YES but if c is NOT a perfect cube (for xample if c=2), then the answer is NO. Not sufficient.

(2) d=\sqrt{c} --> c=d^2. The same here: if c is a perfect square (for example 1^2=1, 2^3=4, 3^2=9, ...), then the answer is YES but if c is NOT a perfect square (for xample if c=2), then the answer is NO. Not sufficient.

(1)+(2) From above we have that d^3=d^2 --> d^3-d^2=0 --> d^2(d-1)=0 --> d=0 (not a valid solution since we are told that d must be positive) or d=1=integer. Sufficient.

Answer: C.

Hope it's clear.
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Re: If c and d are positive, is d an integer ?   [#permalink] 23 Jul 2012, 00:50
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