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# 2^P)*(3^Q) is the greatest positive divisor of 7!*8!/9!.

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Senior Manager
Joined: 30 Aug 2003
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Location: dallas , tx
2^P)*(3^Q) is the greatest positive divisor of 7!*8!/9!. [#permalink]

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16 Jan 2004, 10:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2^P)*(3^Q) is the greatest positive divisor of 7!*8!/9!. Find the product PQ.
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shubhangi

Senior Manager
Joined: 30 Aug 2003
Posts: 322
Location: dallas , tx

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16 Jan 2004, 11:29
working?
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shubhangi

Director
Joined: 28 Oct 2003
Posts: 501
Location: 55405

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16 Jan 2004, 12:20
2^P)*(3^Q) is the greatest positive divisor of 7!*8!/9!.

7!*8!/9!=

1*2*3*4*5*6*7/9

1*2*3*2*2*5*2*3*7/3*3

1*2*2*2*5*2*7/1

2^4, 3^0

4
GMAT Club Legend
Joined: 15 Dec 2003
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16 Jan 2004, 12:28
Agree with you Stoolfi. I also got p=4 q=0. Just a slight detail about the question asking for the product of PQ, not the sum. So answer should be 0?
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Best Regards,

Paul

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Joined: 28 Oct 2003
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Location: 55405

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16 Jan 2004, 12:33
I think you are right, Paul...
16 Jan 2004, 12:33
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