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# PS: Perfect Square

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VP
Joined: 21 Jul 2006
Posts: 1495

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17 Sep 2008, 04:17
If $$x$$ is a perfect square and $$x=(p^a)(q^b)(r^c)(s^d)$$, are prime numbers $$p$$, $$q$$, $$r$$, and $$s$$ distinct?

(1) 18 is a factor of $$ab$$ and $$cd$$

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

Thanks
SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

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17 Sep 2008, 07:01
tarek99 wrote:
If $$x$$ is a perfect square and $$x=(p^a)(q^b)(r^c)(s^d)$$, are prime numbers $$p$$, $$q$$, $$r$$, and $$s$$ distinct?

(1) 18 is a factor of $$ab$$ and $$cd$$

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

Thanks

1)

when a=2k b=2l c=2m d=2n ..

if all a , b, c, d are multiple of 2 then clearly.. p,q,r,s can be distinct or not.

ab=18*i = say i=2
ab=cd= 2*k 2*l = 18*2 =
insuffcient

2)
when a=2k b=l c=2m d=n ..
here b and n not multiple of 2..

So.. q, s must be same to make l+n= multiple of 2.

They are not distinct.

B
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VP
Joined: 21 Jul 2006
Posts: 1495

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17 Sep 2008, 18:25
x2suresh wrote:
tarek99 wrote:
If $$x$$ is a perfect square and $$x=(p^a)(q^b)(r^c)(s^d)$$, are prime numbers $$p$$, $$q$$, $$r$$, and $$s$$ distinct?

(1) 18 is a factor of $$ab$$ and $$cd$$

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

Thanks

1)

when a=2k b=2l c=2m d=2n ..

if all a , b, c, d are multiple of 2 then clearly.. p,q,r,s can be distinct or not.

ab=18*i = say i=2
ab=cd= 2*k 2*l = 18*2 =
insuffcient

2)
when a=2k b=l c=2m d=n ..
here b and n not multiple of 2..

So.. q, s must be same to make l+n= multiple of 2.

They are not distinct.

B

OA is B. Thanks a lot!
Re: PS: Perfect Square   [#permalink] 17 Sep 2008, 18:25
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# PS: Perfect Square

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