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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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Got it - thanks. Sorry, bit late in the night at my end :-).
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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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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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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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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
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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Thnx for your reply. I really appreciate.
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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 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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enigma123
\(p^a q^b r^c s^d=x\), 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.

Hope it's clear.
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enigma123
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.

Bunuel please advise if a,b,c,d are not even then x cannot be a perfect square because square should have primes raised to even powers

Posted from my mobile device
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enigma123
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.

Bunuel please advise if a,b,c,d are not even then x cannot be a perfect square because square should have primes raised to even powers

Posted from my mobile device

Yes. For example, \(2^4*3^4\) IS a perfect square (the square of an integer): \(2^4*3^4 = (2^2*3^2)^2\) but \(2^7*3^4\) is NOT.
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enigma123
\(p^a q^b r^c s^d=x\), 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.


Aren’t we missing the case when either a or b equals 0 and either of c or d equals 0?
In that case, x can be a perfect square without p,q,r,s being distinct
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Bunuel
enigma123
\(p^a q^b r^c s^d=x\), 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.


Aren’t we missing the case when either a or b equals 0 and either of c or d equals 0?
In that case, x can be a perfect square without p,q,r,s being distinct

For (1) yes, but (1) is not sufficient even without that case, so it's not necessary to consider the case when unknowns are 0. For (2), none of them can be 0 because 4 IS a factor of 0, thus unknowns being 0 would violate the statement.
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enigma123
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.

\(x = p^a*q^b*r^c*s^d\) is a perfect square means that if p, q, r and s are all distinct prime numbers, then a, b, c and d must be EVEN integers.

Check here WHY: https://anaprep.com/number-properties-f ... ct-square/

(1) 18 is a factor of ab and cd

This doesn't tell us whether a, b, c and d are all even. They may or may not be. 18 is a factor of ab so it is possible that a = 6 and b = 3 and it is also possible that both a and b are 36 each (even). So not sufficient.

(2) 4 is not a factor of ab and cd

Since both a and b need to be even, 4 must be a factor of ab. If 4 is not a factor of ab, it means they are not both even. Then p, q, r and s CANNOT BE all distinct prime numbers. Answer is a definite NO.
Sufficient Alone.

Answer (B)
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enigma123
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

The same concept is tested in this question that we discussed in our recent webinar too: https://youtu.be/ArveiQ52SrM
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Hi Mam,

it is true that perfect Square must have even power of prime factors but how to find out that prime factors and distinct or same even though all prime factors have even powers

in option 2 where as you have mentioned that may be one is not even power in ab and in CD but how to conclude from that prime factors are distinct or same . if any prime factor have odd factor then it should not perfect square

please correct me if i am wrong anywhere

KarishmaB
enigma123
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.

\(x = p^a*q^b*r^c*s^d\) is a perfect square means that if p, q, r and s are all distinct prime numbers, then a, b, c and d must be EVEN integers.

Check here WHY: https://anaprep.com/number-properties-f ... ct-square/

(1) 18 is a factor of ab and cd

This doesn't tell us whether a, b, c and d are all even. They may or may not be. 18 is a factor of ab so it is possible that a = 6 and b = 3 and it is also possible that both a and b are 36 each (even). So not sufficient.

(2) 4 is not a factor of ab and cd

Since both a and b need to be even, 4 must be a factor of ab. If 4 is not a factor of ab, it means they are not both even. Then p, q, r and s CANNOT BE all distinct prime numbers. Answer is a definite NO.
Sufficient Alone.

Answer (B)
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aronbhati
Hi Mam,

it is true that perfect Square must have even power of prime factors but how to find out that prime factors and distinct or same even though all prime factors have even powers

in option 2 where as you have mentioned that may be one is not even power in ab and in CD but how to conclude from that prime factors are distinct or same . if any prime factor have odd factor then it should not perfect square

please correct me if i am wrong anywhere

KarishmaB
enigma123
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.

\(x = p^a*q^b*r^c*s^d\) is a perfect square means that if p, q, r and s are all distinct prime numbers, then a, b, c and d must be EVEN integers.

Check here WHY: https://anaprep.com/number-properties-f ... ct-square/

(1) 18 is a factor of ab and cd

This doesn't tell us whether a, b, c and d are all even. They may or may not be. 18 is a factor of ab so it is possible that a = 6 and b = 3 and it is also possible that both a and b are 36 each (even). So not sufficient.

(2) 4 is not a factor of ab and cd

Since both a and b need to be even, 4 must be a factor of ab. If 4 is not a factor of ab, it means they are not both even. Then p, q, r and s CANNOT BE all distinct prime numbers. Answer is a definite NO.
Sufficient Alone.

Answer (B)



We are given that x is a perfect square. From statement 2 we conclude that a and b both are not even and c and d both are not even. This means p, q, r and s are NOT all distinct. For example, this is a possibility:

\(x = 3^1*5^{36}*3^3*7^{6}\) (p and r are same with odd exponents but x is a perfect square because exponent of 3 is overall 4- an even number)

If p, q, r and s were all distinct, then a, b, c and d had to be even.
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