Behind the Story of 7 Generous Dwarfs

What was the story?

First thing is very clear that Dwarf 7 must have 0 ounces of milk at the start and end. Let's assume that 'a' be the maximum amount of milk (in ounces) that any dwarf has in his mug at any point of time. 

For a moment, let's assume Dwarf 1 himself has this 'a' amount of milk.

Now, when D1 distributes his 'a' amount of milk among 6 others, D7 receives 'a/6' amount of milk. At this point of time somebody else will be having maximum amount of milk 'a'. Let D2 be that person now having milk 'a'.

Next is D2's turn where he gives a/6 to all. So now D1 has a/6, D7 has 2a/6 and somebody else say D3 has maximum a. Continuing in this way, for each Dwarf's turn gives - 

Now, when we assumed D2 has maximum milk amount a after receiving a/6 from D1, then it's clear that he must had earlier 5a/6. Similarly, D3 had maximum amount of milk a after receiving a/6 from D1 and D2 indicates that he had 4a/6 milk initially. Continuing in this way, we can find the amount of milk each had initially like below.

So at any point of time, the milk distribution is like a, 5a/6, 4a/6, 3a/6, 2a/6, a/6, 0 where amounts are distributed among 7 dwarfs in cyclic order. But the total amount of milk available is 42 ounces.Hence, 

a + 5a/6 + 4a/6 + 3a/6 + 2a/6 + a/6 +  0 = 42

a + 15a/6 = 42

a + 5a/2 = 42

7a = 84

a = 12.

That means at any point of time the maximum amount of milk that one dwarf can have is equal to 12 ounces. And then others would have 10, 8, 6, 4, 2, 0.

To conclude, the 7 dwarfs had 12, 10, 8, 6, 4, 2, 0 ounces of milk initially.