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# 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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11 Mar 2008, 12:27
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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

Last edited by marcodonzelli on 13 Mar 2008, 23:52, edited 2 times in total.

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13 Mar 2008, 06:58
p, q, r, s can be either 2 or any other odd prime number.

Statement 1:
Tells us ab and cd is divisible by 18. That means either a is divisible by 9 and b by 2 or vice-versa or either of a & b is multiple of 18.
We take all the three case:
a=9,b=2
=> 9p, 2q for 9p to be perfect square p must be an odd prime and for 2b to be perfect square q must be even. Only even prime is 2, so q = 4.
Same logic for a=2, b=9
Say a=18,b=5
18p, for this number to be square, p must be even and only even prime is 2.
5q, for this number to be square, q must be odd.

We can say that p and q are different but we cannot say that r, & s is also distinct from p & q. So insufficient.

Statement 2:
Tells us although ab & cd is not divisible by 4. But this is hardly any information as they might be divisible by 2 or by 6 or by any other number other than 4 and its multiple. So this alone is insufficient.

Combining both we have
ab & cd divisible by 18 but not divisible by 4. That means ab & cd is divisible by 2 but not by 4.
If you check statement 1's example, I have not taken a or b to be 4 so point proved there still holds and question is still answered that they are all distinct or not. Although we know 2 pairs a, b & c, d are different.

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13 Mar 2008, 20:09
B

1: ab is something like 18n where n can be any integer if n=1 we get they are not distinct. If n=2, they "can" be distinct.
s insufficient

2:
if ab is not a factor of 4 we can't write ir in the form of (2m)*(2n) which means one of p and q is odd.
Same logic as above applies to cd too.
So we have two even integers and two odd integers among a,b,c and d.
Since x is a pefect square in order to make odd exponents even their base must be same.
which means p, q, r and s can not be distinct.
so sufficient.

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14 Mar 2008, 00:01
sreehari wrote:
B

1: ab is something like 18n where n can be any integer if n=1 we get they are not distinct. If n=2, they "can" be distinct.
s insufficient

2:
if ab is not a factor of 4 we can't write ir in the form of (2m)*(2n) which means one of p and q is odd.
Same logic as above applies to cd too.
So we have two even integers and two odd integers among a,b,c and d.
Since x is a pefect square in order to make odd exponents even their base must be same.
which means p, q, r and s can not be distinct.
so sufficient.

OA is B but I cannot catch this"if ab is not a factor of 4 we can't write ir in the form of (2m)*(2n) which means one of p and q is odd"

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14 Mar 2008, 05:21
The key phrase: "are they distinct?" means that each prime number is different from another. It is possible only if all four powers are even, otherwise, at least two prime numbers have to be equal in order to satisfy "perfect square" condition.

Only B restrict possibility of all-even powers.
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14 Mar 2008, 06:33
marcodonzelli wrote:
sreehari wrote:
B

1: ab is something like 18n where n can be any integer if n=1 we get they are not distinct. If n=2, they "can" be distinct.
s insufficient

2:
if ab is not a factor of 4 we can't write ir in the form of (2m)*(2n) which means one of p and q is odd.
Same logic as above applies to cd too.
So we have two even integers and two odd integers among a,b,c and d.
Since x is a pefect square in order to make odd exponents even their base must be same.
which means p, q, r and s can not be distinct.
so sufficient.

OA is B but I cannot catch this"if ab is not a factor of 4 we can't write ir in the form of (2m)*(2n) which means one of p and q is odd"

If both a and b are even integers, a and b each can be written in the form of 2*something. Since we know that ab is not a factor of 4 you can't do it which means one of them must be odd.

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VP
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14 Mar 2008, 07:38
walker wrote:
The key phrase: "are they distinct?" means that each prime number is different from another. It is possible only if all four powers are even, otherwise, at least two prime numbers have to be equal in order to satisfy "perfect square" condition.

Only B restrict possibility of all-even powers.

so let's say a=3 b=5 c=7 and d=9. obviously we wouldn't have a perfect square. in the case i.e. a=3 and b=3, c=5 and d=7, we would have a perfect square only when p and q are equal, p^3*p^3=p^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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02 Nov 2016, 13:50
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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] 02 Nov 2016, 13:50
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