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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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New post 22 Jul 2011, 20:19
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-n-p-q-p-a ... 01475.html
[Reveal] Spoiler: OA

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

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New post 22 Jul 2011, 21:18
tracyyahoo wrote:
if N=p/q, where p and q are nozero integers, is n an integer?

1) N^2 is an integer
2) 2n+4/2 is an integer

why? pls tell me, thank you.

It is said that n=p/q, so n is a rational number. what is rational number?


Tracy - No need of knowing rational numbers and all - It is separate Maths concept.
Is 2) 2n+4/2 or (2n+4)/2
Anyway either case the answer will be D

Let us walk through the thought process

N=p/q, where p,q are integers
1) N^2 is an integer -
Take N^2 as 3, now N=sqr. root(3)/1 --> It does not satisfy our case
Take N^2 as 9, now N=3/1 --> It satisfies our case

The point is our scenario will be satisfied only when N is a perfect square ( 4,9,25,36,49)
When N is not a perfect square, N will never be an integer, so it will not get satisfied.

So (1) is sufficient

2)
If 2n+4/2 is an integer ==> 2n+2 is an integer ==> 2(n+1) is an integer
This means 2(n+1) is an even integer
So cancelling out 2 on both sides
(n+1) is an integer ==> n will be an integer

If (2n+4)/2 is an integer ==> (n+2) is an integer ==> n is an integer

So (2) is sufficient

So the Ans. is D

Let me know if you are not able to follow the steps.
---- Now going to sleep, bye
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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New post 23 Jul 2011, 05:13
Thank you are the detailed explaination. I understand.

At first I saw this question, I didn't consider p/q. I thought it is useless and I did care why question mentions it.

Now I understand that n^2 is an integer should be fit p/q is an integer.

At first I thought, n^2 is integer, 3 is integer then n is not integer, or 4 is integer then n is integer. I didn't put p/q into consider, now I understand the meaning of the question. Thank you.



krishp84 wrote:
tracyyahoo wrote:
if N=p/q, where p and q are nozero integers, is n an integer?

1) N^2 is an integer
2) 2n+4/2 is an integer

why? pls tell me, thank you.

It is said that n=p/q, so n is a rational number. what is rational number?


Tracy - No need of knowing rational numbers and all - It is separate Maths concept.
Is 2) 2n+4/2 or (2n+4)/2
Anyway either case the answer will be D

Let us walk through the thought process

N=p/q, where p,q are integers
1) N^2 is an integer -
Take N^2 as 3, now N=sqr. root(3)/1 --> It does not satisfy our case
Take N^2 as 9, now N=3/1 --> It satisfies our case

The point is our scenario will be satisfied only when N is a perfect square ( 4,9,25,36,49)
When N is not a perfect square, N will never be an integer, so it will not get satisfied.

So (1) is sufficient

2)
If 2n+4/2 is an integer ==> 2n+2 is an integer ==> 2(n+1) is an integer
This means 2(n+1) is an even integer
So cancelling out 2 on both sides
(n+1) is an integer ==> n will be an integer

If (2n+4)/2 is an integer ==> (n+2) is an integer ==> n is an integer

So (2) is sufficient

So the Ans. is D

Let me know if you are not able to follow the steps.
---- Now going to sleep, bye

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

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New post 23 Jul 2011, 05:57
Kudos are the only way to keep us motivated to post.
Give kudos to anyone who helps you on these forums - after it is all your virtual way of saying Thanks.
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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New post 23 Jul 2011, 09:14
Quote:
if N=p/q, where p and q are nozero integers, is n an integer?

1) N^2 is an integer
2) 2n+4/2 is an integer

Another interesting way to crack this problem:
IF [x], THEN [y]
IF [y], THEN [x]
This means [x]=[y]


So applying the same principles to the below problem :

IF n=p/q, where p and q are non-zero integers and n is an integer, THEN n^2 is a perfect square(that is n is an integer)
IF n^2 is a perfect square, THEN n=p/q, where p and q are nonzero integers and n is an integer

When will both these conditions satisfy, when n^2 is a perfect square
So when n^2 is a perfect sq., then n will be an integer.

So (1) is sufficient...

There are 100 ways of solving a problem ===> Choose the one that is easiest for you.
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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New post 23 Jul 2011, 09:46
Did you get a takeaway here ?

if N=p/q, where p and q are nozero integers, and N^2 is an integer then N^2 is a perfect square
if N=p/q, where p and q are nozero integers, and N^2 is an integer then N is an integer

This is the INFERENCE approach applied to Quant. problems

The beauty of applying this to every problem whether Quant/Verbal is - You will recognize a pattern anytime you see a new question that is a variant of the above one / has some parts of the above one.
QUALITY of time spent on reviewing will be productive and you can get more out of 10 questions than solving 20/30 questions.

So start it now(@least mentally), if you have still not started this practice.
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Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]

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New post 27 Nov 2017, 01:12
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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-n-p-q-p-a ... 01475.html

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Re: If n=(p/q) (p and q are nonzero integers), is an integer?   [#permalink] 27 Nov 2017, 01:12
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