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# If n=(p/q) (p and q are nonzero integers), is an integer?

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If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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21 Sep 2010, 21:43
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If n=(p/q) (p and q are nonzero integers), is an integer?

(1) n^2 is an integer.
(2) (2n+4)/2 is an integer.

(C) 2008 GMAT Club
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Re: I cant understand how the OA is??? [#permalink]

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21 Sep 2010, 22:35
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If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> $$\frac{2n+4}{2}=n+2=integer$$ --> $$n=integer$$. Sufficient.

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Re: I cant understand how the OA is??? [#permalink]

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21 Sep 2010, 23:50
OK now i understood the fundamentals about this question. I was thinking what if n=141/100? in that case n^2 will be 2 but n will not be integer.
Thanks bunuel

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Re: I cant understand how the OA is??? [#permalink]

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22 Sep 2010, 18:31
Bunuel wrote:
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> $$\frac{2n+4}{2}=n+2=integer$$ --> $$n=integer$$. Sufficient.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.
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Re: I cant understand how the OA is??? [#permalink]

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22 Sep 2010, 23:37
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BalakumaranP wrote:
Bunuel wrote:
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> $$\frac{2n+4}{2}=n+2=integer$$ --> $$n=integer$$. Sufficient.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.

That's not true. No reduced fraction when squared can equal to an integer.

Or consider this: $$n^2=\frac{p^2}{q^2}=integer$$ --> $$n=\frac{p}{q}=\sqrt{integer}$$ --> now, square root of an integer is either an integer or an irrational number. But it can not be an irrational number as we have that $$\frac{p}{q}=\frac{integer}{integer}=\sqrt{integer}$$ and we know that an irrational number cannot be expressed as a fraction of two integers, so $$\sqrt{integer}=n=integer$$.

Hope it's clear.
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20 Oct 2010, 13:35
Hi,

please, can me somebody explain why is the first the first statement sufficient?
thanks
Jiri

If n=p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. (2n+4)/2 is an integer.

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20 Oct 2010, 13:40
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Merging similar topics. Please refer to the solution above.

Hope it's clear.
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Re: I cant understand how the OA is??? [#permalink]

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20 Oct 2010, 13:54
Originally answered B, but D's clear now.

Thanks Bunuel!
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Re: I cant understand how the OA is??? [#permalink]

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25 Oct 2010, 13:44
Tricky I picked B but makes perfect sense now. Thanks!

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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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10 Jul 2014, 07:59
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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22 Sep 2015, 08:58
Bunuel wrote:
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> $$\frac{2n+4}{2}=n+2=integer$$ --> $$n=integer$$. Sufficient.

Hi
I am having difficulty understanding statement1.
If n^2=100 then n=10
If n^2=3 then n=sqrt(3) --> why cant we express it as p/q
sorry It may be a dumbest question, but need to clarify..Thanks

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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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25 Sep 2015, 21:14
We can not represent irrational numbers as p/q as the irrational numbers are non repeating non terminating decimals and there is no way we can write p/q for such expression.

One more thing: To check if a square of a number is irrational or not, we need to check if its prime factors have "even powers"
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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13 Jun 2016, 00:07
Bunuel wrote:
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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13 Jun 2016, 00:26
crunchboss wrote:
Bunuel wrote:
If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> $$n^2$$ to be an integer $$n$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ can not be reduced fraction, for example $$\frac{2}{3}$$ or $$\frac{11}{3}$$ as in this case $$n^2$$ won't be an integer). But as $$n$$ can be expressed as the ratio of 2 integers, $$n=\frac{p}{q}$$, then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: $$n$$ is an integer. Sufficient.

You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?

For example, if n = 2/1, or 6/3, or 10/10, ....
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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22 Jun 2017, 12:57
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Re: If n=(p/q) (p and q are nonzero integers), is an integer?   [#permalink] 22 Jun 2017, 12:57
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