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.
