Math Olympiad Problem - Milk Distribution

Source: 1977 All Soviet Union Math Olympiad Problem

Problem:
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?

A few other Math Olympiad Problems on the blog:

Math Olympiad Problems
Russian Coins
Moscow Math Olympiad Problems
Math Olympiad Problem : Overlapping Coins
USA Maths Olympiad Problem - 200th Puzzle

Enjoy!

Update (24/12/2012):
Correct answer posted by Akshay Kumar in comments! Detailed solution posted by me.

Comments

  1. Suppose each one has a,b,c,d,e,f,g ounce of milk initially. Everyone will pour their milk into 6 other cups equally
    so a+b+c+d+e+f=6g
    Similarly
    b+c+d+e+f+g=6a
    Therefore a=g
    and similarly all will be equal
    a=b=c=d=e=f=g = 42/7 = 6 ounce

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  2. initial quantity of milk in order: 12 ,10, 8, 6, 4, 2, 0
    total : 42 ounces

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  3. Correct answer by Akshay Kumar. Detailed Solution is as follows:

    Since the final state matches the initial state, we can imagine this process going on continuously. Consider the dwarf whose cup contains the smallest amount of milk just before he begins pouring. Call him D, and call this quantity of milk a ounces. Since D's cup contains the least milk before pouring, then each other dwarf has an equal or greater amount in his own cup before pouring, and thus gives D at least a/6 ounces. This would leave D with at least a ounces in his own cup just before pouring; the only way he can receive precisely a ounces, as we know he does, is if each other dwarf gives him precisely a/6 ounces and no more, i.e., that each cup, just before pouring, contains the same quantity of milk. So after any given pouring the cups contain a, 5/6a, 4/6a, 3/6a, 2/6a, 1/6a, and 0 ounces; if there are 42 ounces in total then this works out to 12, 10, 8, 6, 4, 2, and 0 ounces.

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  4. You can also generalize the solution for any number. But, you need to have n*(n-1) ounces of milk in total. Then, for each such value, answer would be in AP with difference = -2.
    For example: for 4 dwarfs, if you have 12 ounces of milk in total, then answer would be 6, 4, 2, 0.
    For 3 dwarfs and 6 ounces of milk, answer would be 4, 2, 0.
    :-)

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