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If d is a positive integer, is d^1/2 an integer ?

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If d is a positive integer, is d^1/2 an integer ? [#permalink] New post 15 Feb 2011, 10:53
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kamalkicks wrote:
if d is a positive integer, is \sqrt{d} an integer ?

a . \sqrt{9d} is an integer

b. \sqrt{10d} is not an integer.


is oa correct???????? if then please prove....


Yes, it is.

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

Note that as d is a positive integer then \sqrt{d} is either a positive integer or an irrational number. Also note that the question basically asks whether d is a perfect square.

(1) \sqrt{9d} is an integer --> \sqrt{9d}=3*\sqrt{d}=integer --> \sqrt{d}={integer} (as discussed above because d is an integer \sqrt{d} can not equal to \frac{integer}{3}). Sufficient.

(2) \sqrt{10d} is not an integer --> if d=1 then the answer will be YES but if d=2 then the answer will be NO. Not sufficient.

Answer: A.

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[Reveal] Spoiler: OA

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Re: number properties question [#permalink] New post 15 Feb 2011, 10:55
Statement 1:

sqrt{9d} is an integer

=> sqrt{3^2d} is an integer
=> d is a perfect square
=> sqrt{d} is an integer.

Sufficient

Statement 2:
sqrt{10d} is not an integer.
=> sqrt{2 x 5 x d} is not an integer.
=> since 2 and 5 are not perfect squares, d may or maynot be a perfect square.

Thus sqrt{d} may or maynot be an integer.
e.g.
if d = 9 then sqrt{2 x 5 x 9} is not an integer but sqrt{d}is an integer
if d = 2 then sqrt{2 x 5 x 2} is not an integer and sqrt{2}is not an integer
Insufficient

Ans: 'A'
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Re: number properties question [#permalink] New post 16 Feb 2011, 12:15
clearly a. 10 doesn't have a nice square...
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Re: number properties question [#permalink] New post 17 Feb 2011, 00:07
kamalkicks wrote:
if d is a positive integer, is \sqrt{d} an integer ?

a . \sqrt{9d} is an integer

b. \sqrt{10d} is not an integer.


is oa correct???????? if then please prove....


Original answer is correct.

A. \sqrt{9d} is an integer: tells us 3 sqrt d is an integer. therefore sqrt dhas to be an integer. If it is not, we will never get an integer value. Sufficient.

B. \sqrt{10d} is not an integer:

Case 1: Assume D to be 4
Case 2: Assume D to be 6

In both cases, sqrt (10D) will not be an integer (satisfies the condition)

Now in case 1, sqrt D is an integer but in case 2, sqrt D is not an integer. 2 different answers satisfy the condition. Not sufficient.

Answer A.
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Re: if d is a positive integer, is d an integer ? a . 9d is an [#permalink] New post 27 Nov 2011, 23:17
Quote:
Note that as d is a positive integer then \sqrt{d} is either a positive integer or an irrational number. Also note that the question basically asks whether d is a perfect square.

(1) \sqrt{9d} is an integer --> \sqrt{9d}=3*\sqrt{d}=integer --> \sqrt{d}={integer} (as discussed above because d is an integer \sqrt{d} can not equal to \frac{integer}{3}). Sufficient.


Why cant \sqrt{d} not be an irrational number ??
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Re: if d is a positive integer, is d an integer ? a . 9d is an [#permalink] New post 30 Nov 2011, 03:37
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sohrabkalra wrote:
Quote:
Note that as d is a positive integer then \sqrt{d} is either a positive integer or an irrational number. Also note that the question basically asks whether d is a perfect square.

(1) \sqrt{9d} is an integer --> \sqrt{9d}=3*\sqrt{d}=integer --> \sqrt{d}={integer} (as discussed above because d is an integer \sqrt{d} can not equal to \frac{integer}{3}). Sufficient.


Why cant \sqrt{d} not be an irrational number ??


Because you are given that \sqrt{9d} = 3\sqrt{d} is an integer. Is it possible that 3\sqrt{d} is an integer even though \sqrt{d} is an irrational number?
3*irrational number will still be an irrational number. Hence \sqrt{d} cannot be irrational.
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Re: if d is a positive integer, is d an integer ? a . 9d is an [#permalink] New post 26 Dec 2013, 10:48
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Re: if d is a positive integer, is d an integer ? a . 9d is an   [#permalink] 26 Dec 2013, 10:48
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