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


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is oa correct???????? if then please prove....
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A
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If d is a positive integer, is d^1/2 an integer ? 15 Feb 2011, 11:53
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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....
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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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If d is a positive integer, is d^1/2 an integer ? 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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If d is a positive integer, is d^1/2 an integer ? 16 Feb 2011, 13:15
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clearly a. 10 doesn't have a nice square...
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If d is a positive integer, is d^1/2 an integer ? 17 Feb 2011, 01:07
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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....
Show more

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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If d is a positive integer, is d^1/2 an integer ? 28 Nov 2011, 00:17
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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.
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Why cant \sqrt{d} not be an irrational number ??
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If d is a positive integer, is d^1/2 an integer ? 30 Nov 2011, 04:37
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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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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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If d is a positive integer, is d^1/2 an integer ? 18 Apr 2022, 12:26
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The DS statements on the GMAT have some special characteristics. This question suits as a great example on how to use these to your advantage.

One of these characteristics is: The statements never lie!
In other words: The statements cannot contradict each other!

On this particular question we find out that Statement 1 is sufficient, giving us the definite answer: "Yes, \(\sqrt{d}\) is an integer."

When moving onto Statement 2, remember that statements cannot contradict each other! We therefore know that Statement 2 will in some way give us the possibility of \(\sqrt{d}\) being an integer. So we do not need to check this explicitly for Statement 2.
We can move straight to searching for the possibility of \(\sqrt{d}\) NOT being an integer.
If we find ONE way that \(\sqrt{d}\) is not an integer, we know that the statement is insufficient!
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If d is a positive integer, is d^1/2 an integer ? 14 May 2022, 18:37
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KarishmaB
sohrabkalra
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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Hi KarishmaB / Everyone

Please help me with my approach to this question.

Let us say that d=2

1) √9d is an integer :

√9*2 will this result in an integer........answer is no
can √9*4 be an integer..... answer is yes ( here I have assumed d=4)

2) √10d is not an integer.

when d =2

The statement is true √10*2 is not an integer. The same goes with any value of d

So when the question is asking us Y/N and when we know that √10d is not an integer; This could also mean that d is not an integer, because the statement 2) is correct by itself.

Let me know your thoughts and how else I should be looking at this question.
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If d is a positive integer, is d^1/2 an integer ? 14 May 2022, 22:30
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Anvesh99



Hi KarishmaB / Everyone

Please help me with my approach to this question.

Let us say that d=2

1) √9d is an integer :

√9*2 will this result in an integer........answer is no
can √9*4 be an integer..... answer is yes ( here I have assumed d=4)
Show more

Note that d cannot be 2 here because √9d is an integer but √18 is not an integer.
√9d is same as 3 * √d and if this must be an integer then √d must be an integer.

Anvesh99

2) √10d is not an integer.

when d =2

The statement is true √10*2 is not an integer. The same goes with any value of d

So when the question is asking us Y/N and when we know that √10d is not an integer; This could also mean that d is not an integer, because the statement 2) is correct by itself.

Let me know your thoughts and how else I should be looking at this question.
Show more

Here d can be 2 or 4 or many other values. √20 is not an integer. √40 is not an integer etc.
but √d is not an integer when d = 2 but it is an integer when d = 4.
Hence this statement alone is not sufficient.

Answer (A)
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If d is a positive integer, is d^1/2 an integer ? 14 May 2022, 22:52
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Anvesh99



Hi KarishmaB / Everyone

Please help me with my approach to this question.

Let us say that d=2

1) √9d is an integer :

√9*2 will this result in an integer........answer is no
can √9*4 be an integer..... answer is yes ( here I have assumed d=4)

Note that d cannot be 2 here because √9d is an integer but √18 is not an integer.
√9d is same as 3 * √d and if this must be an integer then √d must be an integer.

Anvesh99

2) √10d is not an integer.

when d =2

The statement is true √10*2 is not an integer. The same goes with any value of d

So when the question is asking us Y/N and when we know that √10d is not an integer; This could also mean that d is not an integer, because the statement 2) is correct by itself.

Let me know your thoughts and how else I should be looking at this question.

Here d can be 2 or 4 or many other values. √20 is not an integer. √40 is not an integer etc.
but √d is not an integer when d = 2 but it is an integer when d = 4.
Hence this statement alone is not sufficient.

Answer (A)
Show more


Thanks KarishmaB :)
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If d is a positive integer, is d^1/2 an integer ? 03 Jul 2025, 10:59
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