Consider the following $6$x$6$ array $$ \begin{matrix} 2 & 0 & 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 1 & 2 & 0 \\ 1 & 0 & 2 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{matrix} $$ One can choose a $k$x$k$ subarray where $1\lt k\lt 6$ and then add $1$ to all elements. Is it possible to perform this action a finite number of times such that every element in the array is a multiple of three?
The sum of all elements in the array is $26$. For all the elements to be divisible by $3$, even the sum has to be divisible by $3$. Choosing a $2$x$2$ subarray will add a total of $4$ to the total sum. A $3$x$3$ will add $9$, a $4$x$4$ will add $16$ and a $5$x$5$ will add $25$. All except a $3$x$3$ subarray will result in the sum being divisible by three. So we can't show it's impossible like this.
I guess some invariant has to be found here, considering the symmetry of the top left $4$x$4$ subarray. Any hint will be appreciated.
Let's divide the original matrix $$ \begin{matrix} 2 & 0 & 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & 1 & 2 & 0 \\ 1 & 0 & 2 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{matrix} $$
into four $3$x$3$ matrices: $$ \begin{array}{lcr|lcr} 2 & 0 & 1 &0 & 2 & 0 \\ 0 & 2 & 0 & 1 & 2 & 0 \\ 1 & 0 & 2 & 0 & 2 & 0 \\ \hline \\ 0 & 1 & 0 & 2 & 2 & 0 \\ 1 & 1 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} $$
Then each $3$x$3$ matrix $A_i$ we transform into matrix $B_i=A_i+A_i+A_i$ to get the final result: $$ \begin{matrix} 6 & 0 & 3 & 0 & 6 & 0 \\ 0 & 6 & 0 & 3 & 6 & 0 \\ 3 & 0 & 6 & 0 & 6 & 0 \\ 0 & 3 & 0 & 6 & 6 & 0 \\ 3 & 3 & 3 & 3 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{matrix} $$