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M04-15

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:22
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Difficulty:

55% (hard)

Question Stats:

53% (01:22) correct 47% (01:00) wrong based on 197 sessions

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If $$p$$ and $$q$$ are nonzero integers and $$n=\frac{p}{q}$$, is $$n$$ an integer?

(1) $$n^2$$ is an integer.

(2) $$\frac{2n+4}{2}$$ is an integer.

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15 Sep 2014, 23:22
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Official Solution:

(1) $$n^2$$ is an integer. Now, $$n^2$$ to be an integer $$n$$ must be either an integer or an irrational number (for example: $$\sqrt{3}$$), (note that $$n$$ cannot be a 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 cannot be an irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction $$\frac{a}{b}$$, where $$a$$ and $$b$$ are integers), so only one option is left: $$n$$ is an integer. Sufficient.

(2) $$\frac{2n+4}{2}$$ is an integer. Given: $$\frac{2n+4}{2}=n+2=\text{integer}$$. So, $$n=\text{integer}$$. Sufficient.

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Joined: 20 Mar 2014
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
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10 Jul 2015, 10:05
1
Bunuel wrote:
If $$p$$ and $$q$$ are nonzero integers and $$n=\frac{p}{q}$$, is $$n$$ an integer?

(1) $$n^2$$ is an integer.

(2) $$\frac{2n+4}{2}$$ is an integer.

Good question. I was wondering why do we have n=P/Q with both p,q = non zero integers, but this information is useful to evaluate statement 1.

Per statement 1, $$n^2$$ = integer. Now $$n^2=4$$ or $$n^2 =2$$ but $$n^2$$=not possible as this will give n = $$\sqrt{2}$$ but as n=P/Q (ratio of integers), it can not be equal to $$\sqrt{2}$$. For n = $$\sqrt{2}$$ = $$\sqrt{2}/1$$ ----> this means that P = $$\sqrt{2}$$ and this will violate the information in the question that P is an integer. Thus this statement is sufficient.

Per statement 2, 2n+4 / 2 = integer ---> n+2 = integer ---> n = integer - 2 = another integer. Thus this statement is sufficient as well.

Thus D is the correct answer.
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Joined: 02 Feb 2018
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12 Jun 2018, 17:33
Hi,

What if p=9 and q=5, then n=p/q=9/5=1.8
then n+2 will be 1.8+2=3. In this case n will be 1.8. So how can be 2nd statement be sufficient.

Thank you!
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Joined: 02 Sep 2009
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12 Jun 2018, 20:16
Priyanka2018 wrote:
Hi,

What if p=9 and q=5, then n=p/q=9/5=1.8
then n+2 will be 1.8+2=3. In this case n will be 1.8. So how can be 2nd statement be sufficient.

Thank you!

When picking numbers you should pick them so that they satisfy the stem and the statement you are testing. Does your example satisfy the second statement, which you are trying to test? (2) says that n + 2 = integer. You numbers does not satisfy this statement.
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19 Sep 2018, 16:46
What if n^2=15? n would equal root(15), otherwise stated as root(30/2) which is not an integer but satisfies the conditions. Please help!
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19 Sep 2018, 19:27
boulousa wrote:
What if n^2=15? n would equal root(15), otherwise stated as root(30/2) which is not an integer but satisfies the conditions. Please help!

n is a rational number, it's a ratio of two integers and thus it cannot equal to irrational number such as $$\sqrt{15}$$. $$\sqrt{15}$$ cannot be expressed as the ratio of two integers. BTW this is explained in the solution.
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Re: M04-15 &nbs [#permalink] 19 Sep 2018, 19:27
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