Distribute $27$ balls evenly among three boxes, transferring exactly $n$ balls in the $n$-th move

Question

There are $3$ boxes, namely, $A$, $B$, and $C$. There are $27$ balls in box $C$. You have to make equal the number of balls in each box. At every $n$-th move, you must transfer exactly $n$ balls from one box to another. You cannot transfer balls between box $A$ and $B$.

How many moves are required to equal the number of balls in all the boxes (if possible)?

A) $7\quad$ B) $8\quad$ C) $9\quad$ D) Not possible

My try:

Since we must start from $1$ and bigger numbers like $7$,$8$, $\ldots$ might not be so appropriate to fulfill our question. I tried to make a list of those numbers whose sum or difference are equal to $9$ (as we want $9$ in every box). Here what I have:

$$\begin{align} 1+2+3+4+5-6 &=9 \\ 2+7 &=9 \\ 2+3+4 &=9 \\ \cdots \end{align}$$ But I could not put them together to get an appropriate solution. Can anyone help?

Answer 1

We want $9$ balls in each box which will require at least 18 balls being moved. This is at least $6$ moves as $\sum_{i=1}^6i = 21$. But it will require an even number of balls being moved after the first 18 as we can't move balls between A and B so after the 18 we are essentially just moving balls back and forth to and from C to fit the move requirements. So the minimum is now $7$ moves with $\sum_{i=1}^7i = 28$ balls being moved.

We can do it in $7$ moves by moving $3$ balls to A on moves one and two, $7$ balls to B on moves three and four, $5$ balls from B to C on move five, $6$ balls to A on move six, and $7$ balls to B on move seven. This results in each box having $9$ balls.