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m04, Q 15

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m04, Q 15 [#permalink] New post 29 Oct 2008, 21:59
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If n = \frac{p}{q} (p and q are nonzero integers), is n an integer?

1. n^2 is an integer
2. \frac{2n+4}{2} is an integer

[Reveal] Spoiler: OA
D


My doubt is if n^2 is an integer then n can be non integer also eg Sqrt2
Then how come D is the answer
[Reveal] Spoiler: OA
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Re: m04, Q 15 [#permalink] New post 27 Feb 2011, 16:18
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achellappa wrote:
i really dont understand any explanation

Now P and q are non zero intergers; so if p=1 and q =2; then p/q = 1/2
p2/q2 = 1/4.. So it is not an integer.

So the answer is B.. any1 get my logic?


p=1 and q=2 are not valid choices for statement (1). It says that n^2 is an integer but for these values of p and q we have that n^2=(1/2)^2=1/4 which is not an integer.

Complete solution:

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.

Answer: D.

Discussed here: i-cant-understand-how-the-oa-is-101475.html

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Re: m04, Q 15 [#permalink] New post 22 Jul 2010, 16:48
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rxs0005 wrote:
p and q are red herring here answer is D


\frac{p}{q} is important to know that n is not irrational. \sqrt{2}, \sqrt{3} etc are irrational and cannot be expressed as ratio.

So when it says N sqaure is integer, N will be interger most of the time. N can be \sqrt{2}but it is ruled out by p/q. So A is suficient

B is sufficient by itself

Hence answer is D
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Re: m04, Q 15 [#permalink] New post 30 Oct 2008, 06:31
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ritula 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

My doubt is if n^2 is an integer then n can be non integere also eg Sqrt2
Then how cum D is the answer


Good question. If I am correctly recalling, I faced it before.

The logic goes like this: If p and q are integers, then the square of the value from p/q never becomes an integer.

Let me ask you a counter question: can you get sqrt (2) dividing p by q such that p and q are integers? If yes, its B and vice versa.
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Re: m04, Q 15 [#permalink] New post 22 Jul 2010, 06:07
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a) n=n^2: if n was a decimal,with for example one decimal digit, then in the tenths' place could be:
1: but n^2 would have in the tenths' place 1 -> n^2 not an integer
2: but n^2 would have in the tenths' place 4 -> n^2 not an integer
3: but n^2 would have in the tenths' place 9 -> n^2 not an integer
4: but n^2 would have in the tenths' place 6 -> n^2 not an integer
5: but n^2 would have in the tenths' place 5 -> n^2 not an integer
6: but n^2 would have in the tenths' place 6 -> n^2 not an integer
7: but n^2 would have in the tenths' place 9 -> n^2 not an integer
8: but n^2 would have in the tenths' place 4 -> n^2 not an integer
9: but n^2 would have in the tenths' place 1 -> n^2 not an integer
So, given that n^2 is an integer means that n is an integer (the decimal unit is 0)
b)(2n+4)/2 is an integer->n+2 is an integer-> n is an integer.
So the best answer is d, every statement provides sufficient information by itself
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Re: m04, Q 15 [#permalink] New post 30 Oct 2008, 22:48
oops yes. I got it. Thanks
GMAT TIGER wrote:
ritula 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

My doubt is if n^2 is an integer then n can be non integere also eg Sqrt2
Then how cum D is the answer


Good question. If I am correctly recalling, I faced it before.

The logic goes like this: If p and q are integers, then the square of the value from p/q never becomes an integer.

Let me ask you a counter question: can you get sqrt (2) dividing p by q such that p and q are integers? If yes, its B and vice versa.
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Re: m04, Q 15 [#permalink] New post 22 Jul 2010, 04:38
But what if p = q = 1? or -1, for that matter? Then you get an integer.
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Re: m04, Q 15 [#permalink] New post 22 Jul 2010, 04:41
Or indeed, if p = q or negatives of each other at any value - eg 2 / 2 becomes 4/4 and n is still 1.
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Re: m04, Q 15 [#permalink] New post 27 Jul 2010, 13:36
For A you get:

n^2 = p^2/q^2 = (p*p) / (q*q) is an integer

You end up with a couple of pieces of information
- q*q is a divisor of p*p.
- q*q & p*p by definition are not prime numbers so you can manipulate as per below

If n = p/q is not an integer, then q is not a divisor p
then q is not a divisor of p*p and (p*p/q) is some arbitrary non integer number
then (p*p/q)/ q = (p*p) / (q*q) can't be an integer either which breaks the whole thing

Thus n = p/q needs to be an integer.
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Re: m04, Q 15 [#permalink] New post 27 Feb 2011, 15:38
i really dont understand any explanation

Now P and q are non zero intergers; so if p=1 and q =2; then p/q = 1/2
p2/q2 = 1/4.. So it is not an integer.

So the answer is B.. any1 get my logic?
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Re: m04, Q 15 [#permalink] New post 27 Feb 2011, 15:42
i really dont understand any explanation

Now P and q are non zero intergers; so if p=1 and q =2; then p/q = 1/2
p2/q2 = 1/4.. So it is not an integer.

So the answer is B.. any1 get my logic?
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Re: m04, Q 15 [#permalink] New post 03 Mar 2011, 20:34
good question thanks! What is the source? I want to try more like these ...
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Re: m04, Q 15 [#permalink] New post 04 Mar 2011, 04:30
Quote:
1) n^2 is an integer --> to be an integer must be either an integer or an irrational number (for example: ), (note that can not be reduced fraction, for example or as in this case won't be an integer). But as can be expressed as the ratio of 2 integers, , 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: is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> --> . Sufficient.

Answer: D.


what if q=1?
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Re: m04, Q 15 [#permalink] New post 04 Mar 2011, 05:04
Expert's post
GiorgosAth wrote:
Quote:
1) n^2 is an integer --> to be an integer must be either an integer or an irrational number (for example: ), (note that can not be reduced fraction, for example or as in this case won't be an integer). But as can be expressed as the ratio of 2 integers, , 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: is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> --> . Sufficient.

Answer: D.


what if q=1?


The same. If q=1 then as given that n=\frac{p}{q} then n=p and as also given that p is an integer then n is an integer too.
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Re: m04, Q 15 [#permalink] New post 26 Jul 2011, 05:08
saxenashobhit wrote:
rxs0005 wrote:
p and q are red herring here answer is D


\frac{p}{q} is important to know that n is not irrational. \sqrt{2}, \sqrt{3} etc are irrational and cannot be expressed as ratio.

So when it says N sqaure is integer, N will be interger most of the time. N can be \sqrt{2}but it is ruled out by p/q. So A is suficient

B is sufficient by itself

Hence answer is D


Thank you for the explanation. Didn't know square roots couldn't be expressed as ratios.
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Re: m04, Q 15 [#permalink] New post 26 Jul 2011, 06:04
This is how I solved it. Let me know if I'm wrong.

before we start, we know that P & Q are integers that =/= 0.
also n= \frac{p}{q}

1. statement 1: n^2=\frac{p^2}{q^2}
we know:
    {p} is divisible by {q}
    \frac{p^2}{q^2} is an integer
the only way for {n}not to be true is for {p} or q to be \sqrt{q} or \sqrt{p} but we know that these are integers so statement 1 is sufficient.

2. statement 2: \frac{2n+4}{2} = {n+2} = \frac{p}{q}+2= integer
we know:
    {p} is divisible by {q}
    integer + integer = integer
    \frac{p}{q} is an integer
sufficient
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Re: m04, Q 15 [#permalink] New post 26 Jul 2011, 11:06
Guys the question states that p and q are non-zero integers. So neither can they be fractions nor they can be irrational numbers.

Hence p and q are only integers but cannot be zero.

Given that n = p/q;
from statement n^2 is integer. Hence n^2 = (p/q)^2 is an integer
so p/q has to be an integer as p and q can only be integers in such a way that p is a multiple of q
and hence n is an integer.

Hence sufficient.

second statement is obviously sufficient for n to be integer.
Hence D
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Re: m04, Q 15 [#permalink] New post 30 Jul 2012, 09:33
Good one. The important point is that it is mentioned that p and q are non-zero integers. This helps resolve the statement(i) which says n*2 is an integer.

Concluded D in ~1 mins and felt I deduced the answer pretty quickly so rechecked. :)
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Re: m04, Q 15 [#permalink] New post 30 Jul 2013, 09:43
Definition of a rational number: Any number of that can be expressed in the form of p/q, where p,q are integers and q !=0.

Definition of Irrational number: a number that cannot be expressed as a rational number. E.g sqrt(7), sqrt(5), etc.,

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Re: m04, Q 15 [#permalink] New post 09 Aug 2013, 03:42
raphaeldzanie1988 wrote:
Im confused with the explanation given.


Okay, I am not a math genius by any means, but this is how it goes:

n = p/q, where we know that p and q are non-zero INTEGERS means that the result (n) will either be an integer or a fraction. It cannot be \sqrt{2} or \sqrt{3}.

So, once you've determined that much, when you go down to the first statement, you are told n^2 is an integer.

Remember, a squared fraction is smaller than the fraction you were squaring.

So:
THE SQUARE OF A FRACTION CANNOT BE AN INTEGER!!! IT WILL BE A SMALLER FRACTION. Therefore, for the square of n to be an integer, p/q must produce a non-zero integer. It could be 1 or it could be 1000, but n would have to be an integer.

Does that explain it?
Re: m04, Q 15   [#permalink] 09 Aug 2013, 03:42
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