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m04, Q 15 [#permalink]
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29 Oct 2008, 22:59
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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 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



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Re: m04, Q 15 [#permalink]
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27 Feb 2011, 17:18
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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: icantunderstandhowtheoais101475.htmlSimilar questions: ifxisapositiveintegerissqrtxaninteger88994.htmlvalueofx107195.htmlnumberpropds106886.htmlnumbersystem106606.htmloddvseventrickquestion106562.htmlquantreview2ndeditionds104421.htmlalgebrads101464.htmlquantreview2ndeditionds104421.htmlisogquantquestionanswerwrong90619.htmlq31og12ds101918.htmlairthmeticds108287.htmlHope it helps.
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Re: m04, Q 15 [#permalink]
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22 Jul 2010, 17:48
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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]
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30 Oct 2008, 07:31
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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]
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22 Jul 2010, 07:07
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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]
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27 Jul 2010, 14:36
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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]
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30 Oct 2008, 23:48
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]
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22 Jul 2010, 05:38
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]
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22 Jul 2010, 05:41
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]
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27 Feb 2011, 16:38
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]
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27 Feb 2011, 16:42
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]
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03 Mar 2011, 21:34
good question thanks! What is the source? I want to try more like these ...



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Re: m04, Q 15 [#permalink]
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04 Mar 2011, 05:30
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]
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04 Mar 2011, 06:04
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]
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26 Jul 2011, 06:08
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]
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26 Jul 2011, 07:04
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]
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26 Jul 2011, 12:06
Guys the question states that p and q are nonzero 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]
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30 Jul 2012, 10:33
Good one. The important point is that it is mentioned that p and q are nonzero 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]
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30 Jul 2013, 10:43
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]
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09 Aug 2013, 04:42
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 nonzero 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 nonzero integer. It could be 1 or it could be 1000, but n would have to be an integer. Does that explain it?







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