p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r

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p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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28 Jan 2012, 00:00

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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.

[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Jan 2012, 02:34, edited 1 time in total.

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Re: Distinct Prime Integers [#permalink]

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28 Jan 2012, 02:31

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enigma123 wrote:

\(p^a q^b r^c s^d=x\), 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.

Answer: B.

Hope it's clear.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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28 Jan 2012, 16:26

Thanks Bunuel - I understand everything apart from this in statement 2:

p,q,r,and s are NOT distinct primes. How???
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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28 Jan 2012, 16:33

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enigma123 wrote:

Thanks Bunuel - I understand everything apart from this in statement 2:

p,q,r,and s are NOT distinct primes. How???

As discussed: 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.

Hope it's clear.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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28 Jan 2012, 16:37

Got it - thanks. Sorry, bit late in the night at my end :-)

.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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29 Jan 2012, 11:26

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!

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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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29 Jan 2012, 17:47

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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 [#permalink]

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30 Jan 2012, 03:39

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!

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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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30 Jan 2012, 13:31

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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.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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30 Jan 2012, 13:43

Thnx for your reply. I really appreciate.

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Re: Distinct Prime Integers [#permalink]

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12 Mar 2012, 11:31

Bunuel wrote:

enigma123 wrote:

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
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Re: Distinct Prime Integers [#permalink]

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12 Mar 2012, 11:49

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kotela wrote:

Bunuel wrote:

enigma123 wrote:

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.

Hope it's clear.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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27 May 2013, 05:30

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Bumping for review and further discussion.
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Re: Distinct Prime Integers [#permalink]

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18 Jul 2013, 06:17

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Bunuel wrote:

enigma123 wrote:

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
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Re: Distinct Prime Integers [#permalink]

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18 Jul 2013, 07:30

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stne wrote:

Bunuel wrote:

enigma123 wrote:

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 think he meant x=\(2^3*3^6*2^3*3^6\).

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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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18 Jul 2013, 07:38

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if that's the case then it seems fine. Awesome observer, you really did observe well, +1 to you.
Strangely no one noticed it till now

Keep it up! Thanks
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Re: Distinct Prime Integers [#permalink]

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18 Jul 2013, 07:57

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stne wrote:

Bunuel wrote:

enigma123 wrote:

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.

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.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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18 Jul 2013, 08:48

The key to this problem is understanding the fact that for \(p, q, r, s\) to be distinct and\(x\)to be a perfect square:

\(a, b , c , d\) MUST be even

or \(ab , cd\)MUST be a multiple of 4.

If \(ab , cd\) are NOT necessarily a multiple of 4 then we cannot be sure that p,q,r,s are distinct.
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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink]

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09 Feb 2014, 06:17

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 [#permalink]

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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r [#permalink] 26 May 2015, 14:49

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p^aq^br^c*s^d=x, where x is a perfect square. If p, q, r

p^aq^br^c*s^d=x, where x is a perfect square. If p, q, r