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ds - nonzero

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SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York
ds - nonzero [#permalink]

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03 Sep 2008, 21:25
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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

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
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Joined: 24 Jun 2008
Posts: 1346
Re: ds - nonzero [#permalink]

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03 Sep 2008, 21:39
x2suresh wrote:
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

The square root of any positive integer is either an integer or an irrational number. So if n^2 is an integer, and n is rational, n is an integer. The second statement is also clearly sufficient, so D.
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VP
Joined: 05 Jul 2008
Posts: 1378
Re: ds - nonzero [#permalink]

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03 Sep 2008, 21:46
n ^ 2 is integer. Can n be a fraction and n ^ 2 result in integer? In other words if either n is a proper or improper fraction
n ^ 2 cannot be an integer and it will be a fraction as well. Suff

2n+4 / 2 is int means n = Int - 2. n is integer. Suff

OA D??
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Joined: 29 Aug 2007
Posts: 2457
Re: ds - nonzero [#permalink]

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03 Sep 2008, 21:50
x2suresh wrote:
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

1: n could be a fraction such as Sqrt(2) or Sqrt(3) or Sqrt(5) or so on.
if n is a square (4), n is an integer. if n is not a square (2 or 3 or 5 or 7, 8 , 10or 11), then n is not an integer. nsf.

2: (2n+4)/2 = n+2 suff.

if an integer (4) plus n is an integer, then n must be an integer....

B..
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VP
Joined: 05 Jul 2008
Posts: 1378
Re: ds - nonzero [#permalink]

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03 Sep 2008, 21:54
GMAT TIGER wrote:
x2suresh wrote:
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

1: n could be a fraction such as Sqrt(2) or Sqrt(3) or Sqrt(5) or so on.
if n is a square (4), n is an integer. if n is not a square (2 or 3 or 5 or 7, 8 , 10or 11), then n is not an integer. nsf.

2: (2n+4)/2 = n+2 suff.

if an integer (4) plus n is an integer, then n must be an integer....

B..

I got lucky even though I did not consider the irrational part but n is clearly mentioned as p/q. Any number that can be expressed as p/q is rational and numbers such as root 2 and 3 etc are irrational

http://en.wikipedia.org/wiki/Irrational_number
Manager
Joined: 09 Jul 2007
Posts: 241
Re: ds - nonzero [#permalink]

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03 Sep 2008, 22:15
n=p/q => n^2=> p/q*p/q= integer; i.e. p/q has to be an integer.
Fyi.. n^2 can not be 3,5 etc as v(3) or v(5) can not be expressed as p/q; where p,q=integer

2nd stement is n+2 is integer=> n is integer

so, IMO D.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346
Re: ds - nonzero [#permalink]

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04 Sep 2008, 05:30
Just to demonstrate $$\sqrt{2}$$ cannot be written as a fraction p/q, where p and q are integers:

If we can write $$\sqrt{2}$$ as a fraction, we can write it as a completely reduced fraction. So suppose $$\sqrt{2} = \frac{p}{q}$$ where $$p$$ and $$q$$ are integers, and $$\frac{p}{q}$$ is a completely reduced fraction. That is, the GCD of p and q is 1. In particular, if we can write $$\sqrt{2}$$ as a fraction p/q of two integers, we must be able to write it as a fraction where p and q are not both even .

$$\sqrt{2} = \frac{p}{q} \\ 2 = \frac{p^2}{q^2} \\ 2q^2 = p^2$$

Since the left side is even, the right side must be even, so $$p^2$$ must be even, and $$p$$ must be even. But if $$p$$ is even, $$p^2$$ must be divisible by $$2^2 = 4$$. If the right side of the equation is divisible by 4, so must be the left side of the equation: $$2q^2$$ must be divisible by 4, which means $$q$$ must be divisible by 2. But we've just shown that $$p$$ and $$q$$ must both be even, which means $$\frac{p}{q}$$ was not a completely reduced fraction, a contradiction. So it cannot be possible to write $$\sqrt{2}$$ as a fraction of two integers, and $$\sqrt{2}$$ must be an irrational number.
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Re: ds - nonzero   [#permalink] 04 Sep 2008, 05:30
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