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Drinking Dwarfs (Posted on 2013-10-23) Difficulty: 3 of 5
Seven dwarfs are sitting at a round table.
Each has a cup, and some cups contain milk.
Each dwarf in turn pours all his milk into the other six cups, dividing it equally among them.

After the seventh dwarf has done this, they find that each cup again contains its initial quantity of milk.

How much milk does each cup contain, if there were 42 ounces of milk altogether?

Source: 1977 All Soviet Union Math Olympiad Problem

No Solution Yet Submitted by Ady TZIDON    
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Solution solution (Two for the price of one.) (spoiler) | Comment 3 of 6 |

This is another problem I solved by misreading the question, as I thought that what was sought was that the dwarfs divide the milk equally between all cups. Call the initial amounts in each cup a,b,c,d,e,f,g. This leads to a complicated distribution chart a/7, b+a/7, c+a/7, the first row after the first move, all the way to (g+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7+(d+a/7+ (b+a/7)/7+(c+a/7+(b+a/7)/7)/7)/7+(e+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7+(d+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7)/7)/7+(f+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7+(d+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7)/7+(e+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7+(d+a/7+(b+a/7)/7+(c+a/7+(b+a/7)/7)/7)/7)/7)/7)/7 in the bottom right hand corner of the final row after the last move, calling for some lateral thinking.

Call the first dwarf Arnie, the second Beefy, and the last two Feeble and Gloomy (for reasons that will become obvious). Call the complicated expressions after each move A1, B1,...G7.

After the last move, Gloomy's cup contains just the same as it did at the start. So each of Gloomy's contributions in G7, which are the only contributions he makes, must be the same as what he had at the outset. Instead of G7, we can just write g. But then what Feeble contributed in the penultimate move must have been what he had at the start, less what he received in the last move, i.e.(f-g). And we can do this replacing all the complicated expressions for each dwarf and each move until Arnie's column for example reads a, a/7, (b-c) (c-d), (d-e),(e-f),(f-g),g. In other words, a/7+(b-c)+(c-d)+(d-e)+(e-f)+(f-g)+g=a; cancelling and substituting, b=6a/7, c=5a/7, etc.

Since there were 42 units of milk to start with, the proportions would be{10.5,9,7.5,6,4.5,3,1.5.}

BUT: The division is not between all cups, but between 'the other 6 cups'. We can see from the above solution to the related problem that immediately before each move, the amount in the current dwarf's cup is equal to what was in Archie's cup to start with. Now we have a/6+b=a for distribution in Beefy's cup after the first move. b=5a/6, c=4a/6,..., g=0a/6. Gloomy starts and finishes with no milk at all, hence his name.

The solution to the actual problem is therefore {12,10,8,6,4,2,0}

A neat puzzle.

  Posted by broll on 2013-10-24 01:22:09
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