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\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we cannot get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*3^6*2^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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29 Jan 2012, 10:26
A silly question! If you loose on such a question, what score one should expect for quant? (By the way I understood what you explained here, Bunuel, but you would be there on exam day, right!
Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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29 Jan 2012, 16:47
1
docabuzar wrote:
A silly question! If you loose on such a question, what score one should expect for quant? (By the way I understood what you explained here, Bunuel, but you would be there on exam day, right!
Or this is a silly post that adds no value. On the GMAT, only the experimental ones are silly. This is a pretty tough question. If you think it is an easy one, you should consider providing your own explanation - you will learn quite a bit when you try to teach someone.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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30 Jan 2012, 02:39
You mis understood my statement. Or may be I wrote it in a wrong way.
What I meant was that I want to ask a silly question, i.e., if someone looses on such a question in GMAT what score he/she should expect. (Acutally I do understand bunuel's explanation to this Q, but I donot think that I will be able to re-produce the concept if this Q appears with some varaition in GMAT)
As far as this question is concerned, for me, it was very tough. I have always appreciated the knowledge bunuel (& for that matter anyone) expresses here in these forum esp the ease with which bunuel explains so many twists in a single question. I m still learning. Cheers!
Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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30 Jan 2012, 12:31
1
docabuzar wrote:
You mis understood my statement. Or may be I wrote it in a wrong way.
What I meant was that I want to ask a silly question, ...
Oh, don't worry, no hard feelings whatsoever.
docabuzar wrote:
If someone looses on such a question in GMAT what score he/she should expect. (Acutally I do understand bunuel's explanation to this Q, but I donot think that I will be able to re-produce the concept if this Q appears with some varaition in GMAT)
As far as this question is concerned, for me, it was very tough. I have always appreciated the knowledge bunuel (& for that matter anyone) expresses here in these forum esp the ease with which bunuel explains so many twists in a single question. I m still learning. Cheers!
You are right, it's a quite hard question, probably 700+. So if one answers incorrectly to 1 or 2 of such questions he/she can still expect a pretty decent score.
_________________
\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we can not get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Answer: B.
Hope it's clear.
Hi Bunnel
From statement 2 how did you say that P Q R S are distinct primes as we have information only on a,b,c,d???
\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we can not get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Answer: B.
Hope it's clear.
Hi Bunnel
From statement 2 how did you say that P Q R S are distinct primes as we have information only on a,b,c,d???
Thanks in advance
Well, I'm saying exactly the opposite for (2): \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes.
As for the connection between a, b, c, d and p, q, r, s: if \(p\), \(q\), \(r\), and \(s\) are distinct primes, then \(a\), \(b\), \(c\), and \(d\) MUST be even (all of them).
From (2) we get that NOT all from \(a\), \(b\), \(c\), and \(d\) are even, hence \(p\), \(q\), \(r\), and \(s\) are NOT distinct.
\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we can not get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Answer: B.
Hope it's clear.
I have a doubt regarding this, above we have taken statement 1 has 2 cases first case : \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes.
Now how can we take this case? because it says that x is a perfect square so ultimately total powers of x should be odd, but in this case total powers of x is even , hence x is not a perfect square which contradicts what is given, so shouldn't this case be invalid. second case for statement 1 of course looks fine.
shouldn't our objective be to find x such that x is a perfect square and p q r s are not distinct and 18 is a factor of ab and cd if we can find such a case then we will have two cases of perfect square: one with distinct p q r s and other with non distinct p q r s hence insufficient
but here the first case for statement 1 x is not a perfect square so how can we take this as a valid case ,or am I missing something, can anyone help?
\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we can not get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Answer: B.
Hope it's clear.
I have a doubt regarding this, above we have taken statement 1 has 2 cases first case : \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes.
Now how can we take this case? because it says that x is a perfect square so ultimately total powers of x should be odd, but in this case total powers of x is even , hence x is not a perfect square which contradicts what is given, so shouldn't this case be invalid. second case for statement 1 of course looks fine.
shouldn't our objective be to find x such that x is a perfect square and p q r s are not distinct and 18 is a factor of ab and cd if we can find such a case then we will have two cases of perfect square: one with distinct p q r s and other with non distinct p q r s hence insufficient
but here the first case for statement 1 x is not a perfect square so how can we take this as a valid case ,or am I missing something, can anyone help?
\(p^a q^b r^c s^d=x\), where x is a perfect square. If p, q, r, and s are prime integers, are they distinct? (1) 18 is a factor of ab and cd (2) 4 is not a factor of ab and cd
Any idea how to solve this question please? I don't have an OA unfortunately.
p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?
First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.
(1) 18 is a factor of ab and cd --> we can not get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.
(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.
Answer: B.
Hope it's clear.
I have a doubt regarding this, above we have taken statement 1 has 2 cases first case : \(p^a*q^b*r^c*s^d=2^3*2^6*3^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes.
Now how can we take this case? because it says that x is a perfect square so ultimately total powers of x should be odd, but in this case total powers of x is even , hence x is not a perfect square which contradicts what is given, so shouldn't this case be invalid. second case for statement 1 of course looks fine.
shouldn't our objective be to find x such that x is a perfect square and p q r s are not distinct and 18 is a factor of ab and cd if we can find such a case then we will have two cases of perfect square: one with distinct p q r s and other with non distinct p q r s hence insufficient
but here the first case for statement 1 x is not a perfect square so how can we take this as a valid case ,or am I missing something, can anyone help?
Thanks
It was a typo. Edited. Should have been: \(p^a*q^b*r^c*s^d=2^3*3^6*2^3*3^6\). Thank you. +1.
_________________
Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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09 Feb 2014, 05:17
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Very good question. Thanks for posting. I chose the wrong answer. BUT If p,q,r & s are distinct primes then a,b,c,d each should be even. if one of them is odd (let's say a) then at least one more of them must also be odd and two of the primes must be equal otherwise the value x cannot be a perfect square. Knowing if one of a,b,c & d is odd answers the question. Hence B. I hope I made sense.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r
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