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#1 2008-09-16 04:56:27
- Kurre
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- Registered: 2006-07-18
- Posts: 280
Hourglasses
Given two hourglasses, on a respectively b minutes, can we always measure c minutes? If not, what criteria must hold for it to be possible?
Last edited by Kurre (2008-09-16 04:56:36)
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#2 2008-09-16 07:11:27
- mathsyperson
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- Registered: 2005-06-22
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Re: Hourglasses
The first one is that c needs to be a multiple of hcf(a,b), since neither of the hourglasses can work with increments smaller than that. Also, c needs to be no less than min(a,b).
I'm pretty sure that if c > ab and it meets the hcf requirement, then it's definitely timeable.
Times between min(a,b) and ab are a bit trickier to think about though.
Why did the vector cross the road?
It wanted to be normal.
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#3 2008-09-16 07:31:30
- Kurre
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Re: Hourglasses
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