The Dwarven Drinking Puzzle

The Dwarven Drinking Puzzle

This is one of my favourite puzzles, courtesy of the ever brilliant Lewis Carroll. You are one of seven dwarves sitting around a great stone table in your mountain hall. As usual you are drinking mead from tankards after a hard day in the mines and you start a rather odd custom. Beginning with you, you divide your mead equally into the other six tankards leaving you with nothing and everyone else slightly topped up. Next the dwarf to your left does the same; dividing their current amount equally between the other dwarves including you. This same action continues around the table until all seven of you have gone through the motions.


When everyone has had a go, you realise that everyone has exactly the same amount of mead in their tankard as when they started. Given that there is exactly 42 fluid ounces of mead in total, how much did each dwarf have at the start?

I encourage you to have a go. 3 hints and an answer below.

 

 

 

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Hint 1, how much mead did dwarf number 7 have at the start?

 


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Hint 2, if you go into fractions in your working you will never escape them.

 


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Hint 3, if you ignore the 42 total, can you find any set of numbers that work? What do they sum to?

 

 

 

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Answer. Ok well let's start by looking at the hints. The last dwarf emptied their tankard just before the end, so since they ended with the same amount that they started with, they must have had nothing to start with.

The only drink that the penultimate dwarf receives is a single amount from the last dwarf right at the end; similarly the fifth dwarf receives two amounts before the end and so on. This suggests that the distribution is a decreasing sequence starting from you and ending at zero.

If we think about sensible amounts that you could have started with, since you need to distribute into six equal parts, you must have had a multiple of six fluid ounces originally; otherwise we are going to end up working in sixths after your pour, then thirty sixths and so on into increasingly atrocious fractions.

So let's say you have 6 to start with so you will give 1 to everyone. But now dwarf number 2 needs to have a multiple of 6, so they must have had 1 fewer than a multiple of 6 to start with and we know it is a smaller amount, so they must have had 5.

Carrying on this line of logic we get that the seven dwarfs had [6,5,4,3,2,1,0] to start with and after each pouring the distribution just rotates around the table, so [0,6,5,4,3,2,1], then [1,0,6,5,4,3,2] and so on until just before the final pouring we have [5,4,3,2,1,0,6]. This is a solution which neatly rotates around, but summing it up it gives only 21 fluid ounces.

By doubling the quantities involved we get the 42 fluid ounces required and so the final answer is that the dwarves have [12,10,8,6,4,2,0] in their tankards to start.

If you are sat at a table with a whole lot of mathematicians (which hopefully is a common occurrence in your life), then you can change the number 42 into n^2-n for n people and the problem will have the same structure.

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