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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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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.

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[Reveal] Spoiler: OA

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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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16 Feb 2011, 13:15
clearly a. 10 doesn't have a nice square...
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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....

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.

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Re: if d is a positive integer, is d an integer ? a . 9d is an [#permalink]

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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 an integer ? a . 9d is an [#permalink]

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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 an integer ? a . 9d is an [#permalink]

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26 Dec 2013, 11:48
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Re: If d is a positive integer, is d^1/2 an integer ? [#permalink]

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22 Mar 2016, 07:05
Hello from the GMAT Club BumpBot!

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Re: If d is a positive integer, is d^1/2 an integer ?   [#permalink] 22 Mar 2016, 07:05
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