Question 98
Let $n \geq 3$ be a fixed natural number. There are $n$ boxes $A_{1}, A_{2}, \ldots, A_{n}$, each with a number of stones in it, $a_{1}, a_{2}, \ldots, a_{n}$, such that $a_{1}+a_{2}+\cdots+a_{n}=3 n$. A move consists of the following actions:
choose a box and distribute all stones in the box over the $n$ boxes (including the chosen box) such that for every two boxes the numbers of added stones differ by at most 1.
For a distribution $a_{1}, a_{2}, \ldots, a_{n}$ we define $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ as the least number of moves needed to get all stones into one box. Let $M_{n}$ be the maximum of $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ over all possible distributions $a_{1}, a_{2}, \ldots, a_{n}$ with $a_{1}+a_{2}+\cdots+a_{n}=3 n$. Determine $M_{n}$ and all distributions $a_{1}, a_{2}, \ldots, a_{n}$ for which $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)=M_{n}$.
Example. If $n=4$ and the boxes contain 2, 6, 0, 4 stones, then we can distribute the 2 stones from box $A_{1}$ as 1,0,1,0. After this move the numbers of stones per box are 1, 6, 1, 4.
choose a box and distribute all stones in the box over the $n$ boxes (including the chosen box) such that for every two boxes the numbers of added stones differ by at most 1.
For a distribution $a_{1}, a_{2}, \ldots, a_{n}$ we define $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ as the least number of moves needed to get all stones into one box. Let $M_{n}$ be the maximum of $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ over all possible distributions $a_{1}, a_{2}, \ldots, a_{n}$ with $a_{1}+a_{2}+\cdots+a_{n}=3 n$. Determine $M_{n}$ and all distributions $a_{1}, a_{2}, \ldots, a_{n}$ for which $f\left(a_{1}, a_{2}, \ldots, a_{n}\right)=M_{n}$.
Example. If $n=4$ and the boxes contain 2, 6, 0, 4 stones, then we can distribute the 2 stones from box $A_{1}$ as 1,0,1,0. After this move the numbers of stones per box are 1, 6, 1, 4.