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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]

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New post 04 Apr 2010, 12:46
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If d is a positive integer, is \(\sqrt{d}\) an integer ?

(1) \(\sqrt{9d}\) is an integer

(2) \(\sqrt{10d}\) is not an integer.


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

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

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New post 15 Feb 2011, 11:53
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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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Re: If d is a positive integer, is d^1/2 an integer ?  [#permalink]

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New post 15 Feb 2011, 11:55
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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: If d is a positive integer, is d^1/2 an integer ?  [#permalink]

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New post 16 Feb 2011, 13:15
clearly a. 10 doesn't have a nice square...
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Re: If d is a positive integer, is d^1/2 an integer ?  [#permalink]

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New post 17 Feb 2011, 01:07
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 \(sqrt d\)has 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^1/2 an integer ?  [#permalink]

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New post 28 Nov 2011, 00:17
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^1/2 an integer ?  [#permalink]

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New post 30 Nov 2011, 04:37
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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^1/2 an integer ?  [#permalink]

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Re: If d is a positive integer, is d^1/2 an integer ?   [#permalink] 07 Apr 2019, 00:47
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