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

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

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New post 28 Oct 2009, 03:17
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A
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D
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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
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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 03:39

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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 08:01
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I think it is D...Here is the explanation....
If n=p/q ( p and q are nonzero integers), is n an integer?

(1) n^2 is an integer
Condition says n2 is integer....means n2 has to be a positive square number (square cannot be negative).... 4, 9, 16....
Why only perfect square because if n is not an integer than n2 cannot be an integer too (but condition 1 says n2 is an integer)... so Sufficient...
(2) (2n+4)/2 is an integer
As per condition 2....(2n+4)/2 is an integer....
it can only be integer ..if n is an integer e.g. -10, -5, 0, 4, 9, 50 etc.... means n has to be integer for this ....(2n+4)/2 is an integer...... so Sufficient...

please correct if my understanding is wrong....

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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 10:31
Hussain15 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


Yes the answer is D, but there is a catch in statement 1 that has been ignored so far.

So for the sake of clarification:

The fact that n^2 is an integer need not imply that n is an integer. If n=sqrt(3) then n^2 = 3 which is an integer. BUT remember we are told in the Q that n=p/q, where p and q are non-zero integers. Since we cannot get a square-root of a number which is not a perfect square by dividing to integers the statement IS sufficient.

Statement 2 is obviously sufficient - If in doubt just plug in numbers.

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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 10:54
D

(A) n=sqrt(x) is integer and given n=p/q [e.g sqrt(2) is irrational, but out of scope due to given info] sufficient
(B) (2n+4)/2 = n+2. sufficient

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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 17:01
IMO D,

andershv
I like your explanation
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Re: Is n an integer [#permalink]

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New post 28 Oct 2009, 19:59
Thanks for the input! Obviously the trick was in point 1.

Answer is "D".

Can we use the concept of prime factorization to determine the sufficiency of statement 1?? I mean some technical way, rather plugging in the numbers.
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Re: Is n an integer [#permalink]

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New post 30 Oct 2009, 18:31
I think, this problem is better solved by not plugging numbers, but by knowing the number properties.
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Re: If n=p/q ( p and q are nonzero integers), is n an integer? [#permalink]

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Re: If n=p/q ( p and q are nonzero integers), is n an integer?   [#permalink] 15 Aug 2017, 04:48
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