mathematics - Winnie-the-Pooh and the 27 honey pots

Winnie-the-Pooh keeps his $27$ honey pots in the larder. Each pot contains up to $1$ kilogram of honey, and different pots contain different quantities of honey. All $27$ pots together contain $17$ kilogram of honey.

Every day, Winnie-the-Pooh selects $7$ pots, picks a real number $x$, and then eats exactly $x$ kilogram of honey from each of the selected pots.

Question: Is it always (independently of the initial distribution of honey) possible for Winnie to empty all $27$ honey pots in a finite number of days?

Answer

Here is a rigorous proof that Pooh can always succeed.

Let $P_i$ be the amount of honey in the $i^{th}$ pot when sorted in decreasing honey contents, so $1\ge P_1\ge P_2\ge \dots\ge P_{27}\ge0$ and $P_1+\dots+P_{27}=17$.

Using the beginning of Sleafar's solution, we now need only show how to equalize the seven heaviest pots. At any point, let $m$ be the number of pots which have the same amount of honey as the heaviest, and let $n$ be the number of nonempty pots (so initially, $m=1$ and $n=27$). We show that, as long as $m<7$ and $n\ge 14$, it is possible to either increase $m$ or decrease $n$. We then argue why $n$ can't drop below $14$ before $m$ reaches $7$, so that eventually we equalize the top $7$ pots.

Here is the method. Let $x$ be the gap in honey contents between the heaviest $m$ pots and the next strictly lighter pot. Each day for the next $n-7$ days, Pooh eats $\frac{x}{n-7}$ kg from each of the $m$ heaviest pots. Each of these days, he must eat the same amount from some $7-m$ other pots. He does this in such a way that the $n-7$ pots $P_8,P_9,\dots,P_n$ are eaten from equally over this $n-7$ day period (for example, by arranging the nonempty pots in a circle, and eating from each consecutive segment of length $7-m$).

As long as each of the nonempty pots contain at least $\frac{x}{n-7}\cdot (7-m)$ kg of honey, then this will work, causing the $m$ heaviest pots to decrease to the next highest weight, thus increasing the number $m$. If the nonempty pots do not contain enough honey to do this, then adjust $x$ so that the smallest pot contains exactly $\frac{x}{n-7}\cdot (7-m)$. Then doing the procedure will empty that pot, thus decreasing $n$.

Now, why can't $n$ drop below $14$ before $m$ reaches $7$? First off, the amount of honey that needs to be removed from pots 8 through $n$ in order to equalize pots 1 through 7 is equal to $$ 6(P_1-P_2)+5(P_2-P_3)+4(P_3-P_4)+3(P_4-P_5)+2(P_5-P_6)+(P_6-P_7) $$ This is because, during the phase when pots $P_1,\dots,P_m$ are equal, we have to remove a total of $P_{m}-P_{m+1}$ from each of the first $m$ pots, and $(7-m)$ times that from the last pots 8 through $n$. A different way of writing this is $$ 6P_1-P_2-P_3-\dots -P_7\le 6 - 6P_7< 3 $$ The last inequality follows since $P_7> \frac12$. Why is this true? If not, then the first 6 pots would initially contain at most $6$ kg, and the last 21 would contain at most $\frac12$, so that the total amount of honey was at most $6+\frac12\cdot 21=16.5$, which is impossible since we started with $17$ kg of honey.

So, the procedure will remove less than $3$ kg from the last 20 pots. However, the first 14 pots contain at most 14 kg, meaning the lightest pots together contain at least $17-14=3$ kg. Since our procedure eliminates the lightest pots first, the first 14 pots will not be emptied, as claimed.