Is this proposition posible?

Question

In a board, you have $13$ White round pieces, $15$ Black round pieces, and $17$ Red round pieces. In each round you can choose two different color pieces and change them with two other pieces of the remaining color. Doing this every round, it's possible to get all the pieces to be the same color ?

I think is not possible, and I think that it has to be with the case that the number of pieces is separated by $2$. That is, let $3x$ be the number of total pieces divided 3, that is $x=15$. The problem I see is that:

In this case $White=x-2,Black=x,Red=x+2$ $$(\underbrace{x-2}_\text{White},\underbrace{x}_\text{Black},\underbrace{x+2}_\text{Red})$$

In this configuration is not possible to get to $$(x,x,x)$$

From this I can get to all from $1 $ Color $$(3x,0,0)$$

Answer 1

Let us solve a more general problem. Instead of the particular numbers let us assume that there are $w_1$ white chips, $b_1$ black chips and $r_1$ red chips. Now let in the $n$-th step the number of chips become $(w_n,b_n,r_n)$. Then, Notice that for the transformation $(w_n,b_n,r_n) \rightarrow (w_{n+1},b_{n+1},r_{n+1})$ we have one of the three following, $$(w_{n+1},b_{n+1},r_{n+1})=(w_n-1,b_n-1,r_n+2)$$ Or, $$(w_{n+1},b_{n+1},r_{n+1})=(w_n-1,b_n+2,r_n-1)$$ Or, $$(w_{n+1},b_{n+1},r_{n+1})=(w_n+2,b_n-1,r_n-1)$$

Now note that in each case $I=(w_n-b_n) \pmod 3$ is an invariant. Also note that $(b_n-r_n) \equiv 0 \pmod 3$ and $(r_n-w_n) \equiv 0 \pmod 3$ are also invariant. So $I \equiv 0\pmod 3$ combined with $w_1+b_1+r_1\equiv 0 \pmod 3$ is the required condition to reach the monochromatic state.

Answer 2

For a brute force solution for why it isn't possible, we'll use linear algebra. Let: $$ \begin{bmatrix} w \\ b \\ r \end{bmatrix}$$ represent the fact that there are $w$ white, $b$ black, and $r$ red pieces at some round, so that we initially have the vector: $$ \begin{bmatrix} 13 \\ 15 \\ 17 \end{bmatrix} $$ Then each possible move can be encoded by adding one of the following three vectors: $$ \begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix} \qquad\text{or}\qquad \begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix} \qquad\text{or}\qquad \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} $$ Notice that the moves commute (the order in which they are done doesn't matter). So the problem reduces to finding some nonnegative integer scalars $c_1,c_2,c_3 \in \{0, 1, 2, \ldots\}$ such that: $$ \begin{bmatrix} 15 \\ 15 \\ 15 \end{bmatrix} = \begin{bmatrix} 13 \\ 15 \\ 17 \end{bmatrix} + c_1\begin{bmatrix} 2 \\ -1 \\ -1 \end{bmatrix} + c_2\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix} + c_3\begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} $$ Rewriting as the augmented matrix: $$ \left[\begin{array}{ccc|c} 2 & -1 & -1 & 2 \\ -1 & 2 & -1 & 0 \\ -1 & -1 & 2 & -2 \end{array}\right] $$ we may solve this system to obtain solutions of the form: $$ \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = c_3\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + \begin{bmatrix} 4/3\\ 2/3 \\ 0 \end{bmatrix} $$ where $c_3$ is free to be any nonnegative integer we choose. But since no choice of $c_3$ will eliminate the fractions so that $c_1,c_2$ are also integers, it's impossible to get all pieces to be the same colour.