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November 1st, 2009, 07:18 AM   #1
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Confused Problem

How many ways are there to partition a 10 by 12 rectangle into four regions using sides x+a by x+b, where x, a, and b are positive integers?

All I got is (x+10)(x+12) and then FOIL I am not sure if this is in any way correct which is why I am stuck. Any help is appreciated.
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November 1st, 2009, 09:16 AM   #2
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Re: Confused Problem

[quote="Scary69l"]How many ways are there to partition a 10 by 12 rectangle into four regions using sides x+a by x+b, where x, a, and b are positive integers?

All I got is (x+10)(x+12) and then FOIL I am not sure if this is in any way correct which is why I am stuck. Any help is appreciated.[/quote

The point is that BOTH x and a (or b) are integers. So side can be (1, 2, 3, 4...10) or (1, 2, 3...12). It gives you limited number of divisions. Side shows how many, let's called it, degrees of freedom (DOF) we have.
If it is x=2 b=2 and a=1 then in 12 side we have of 8 of 4 (x+b) DOF and 9 od 3 (x+a) DOF. 8 because 12-4=8, in other words we can put side x+b in 8 positions. It is for ANY x or a or b.

However x+a and x+b must be - if I understand properly the problem - divider of 12 or 10, and we have limitation to sum of them. And here I need clarification:
is it that x+a corresponds to side 12 and x+b corresponds to side 10 or should they correspond BOTH?
I need it because solution depends of it.
For example if x+a corresponds to 12 AND x+b corresponds to 10 THEN we have x in both and it is very narrow 'space' for a and b (as it must be x+a divider of 12 and x+b divider of 10).

Please clarify that means conditions.
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November 1st, 2009, 10:43 AM   #3
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Re: Confused Problem

I myself am completely confused with the question. Which is why I will probably skip it for now. Thanks for your help
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