10 zeroes, 9 ones, and 8 twos are written on the board. Given the operations, can you set all numbers to become zero?

Question

You are allowed to replace 0 and 1 with two 2s, 0 and 2 with two 1s, or one and two by two zeroes.
My thinking was that you had to set all numbers to 1 and 2. However, this isn't possible since there are an odd number of numbers and each operation modifies two.

Answer 1

Hint: Consider the total value modulo $3$ of the initial configuration and the final configuration. For each move, what is the effect of that move on the total value modulo $3$ of a configuration?

Answer 2

The modulo 3 answer is very good. Here's a geometrical solution:

These operations can be interpreted as moves in a three-dimensional space. Let $x$ be the number of $0$s, $y$ be the number of $1$s, and $z$ be the number of $2$s. We start out at $\vec{r} = (x,y,z) = (10,9,8)$. The operation "$0$ $1$ becomes $2$ $2$" can be interpreted as adding $\vec{u}=(-1,-1,2)$ to the current position. Likewise, "$0$ $2$ becomes $1$ $1$" and "$1$ $2$ becomes $0$ $0$" can be interpreted as adding $\vec{v} = (-1,2,-1)$ or $\vec{w}=(2,-1,-1)$, respectively, to the current position.

A solution to this problem would consist of three non-negative integers $a$, $b$, and $c$ such that $$ (10,9,8) + a\, \vec{u} + b\, \vec{v} + c\, \vec{w} = (27, 0, 0)\, . $$ Taking the $y$ and $z$ components of this expression yields: \begin{align} 9 - a + 2b - c &= 0\\ 8 + 2a - b - c &= 0 \end{align} Subtracting the second of these from the first results in $$ 3(a-b)=1\, , $$ which has no solution in integers $a$ and $b$. Thus there is no solution to the original problem.