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The problem is this
we have a box that has in it 3 vessels which have the property that they empty their content once they are full. Their content is being emptied to the vessel to their right. Once the most right vessel is full then it just empties its content out of the box.
Here is a picture to illustrate this a bit better.
The vessels at the beggining are empty.
The only lead i have about it is that out of the box i get either 12 glasses of water or 15.

How many glasses of water is the volume of each vessel??

Anivos
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  • What triggers dumping the third one when it isn't full? It sounds like you should always get the amount of the third vessel. – Ross Millikan May 09 '18 at 23:29
  • at first it appears so, but imagine this, you have for example a second vessel that fits 12 glasses and a third that fits 3, then as soon as the second is filled everything is going to come out of the third too, so you will get 12 glasses out – Anivos May 09 '18 at 23:33
  • You need to make the rules clear. What happens if you just have a 9 and a 6 vessel? When the 9 fills it spills to the 6. Does the 6 spill out 6 and keep the other 3 or does 9 spill out? – Ross Millikan May 10 '18 at 14:39
  • yes indeed, when the 9 fills it means then that the 6 at first spills out 6 and then keeps 3. If it was a 9 and a 3 vessel, then once the 9 fills then the 3 just outputs 3 times itself. hope its clear enough – Anivos May 10 '18 at 17:32

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I can do it with just two vessels, $14$ and $3$. When the $14$ spills, the $3$ will dump four times, releasing $12$ and holding $2$. The next time the $14$ spills the $3$ will dump five times, releasing $15$ and holding $1$. The third time the $14$ spills the $3$ will again dump five times, ending up empty, and we go around again.

Any first vessel that is between $12$ and $15$ will work. The fraction of spills that is $15$ will rise from $0$ to $1$ linearly with the capacity of the first vessel. If I have to use a third I can just put anything less than $15$ in front.

Ross Millikan
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