Made an obvious mistake, will be fixed

bin_sherlo wrote: We claim that $B$ wins. Rephrased Problem: There are $675$ times $1$,$675$ times $2$ and $674$ times $0$ on a board. On each turn, $A$ chooses $a,b,c$ and $B$ writes one of $(a+b+c)+\{a,b,c\}$ and erases $a,b,c$. After some turns, there will be two numbers on the board. If sum of the numbers written on the board is a multiple of $3$, then $B$ wins and $A$ wins otherwise. Who wins? We work on modulo $3$. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 0&0&0&\rightarrow&0\\\hline 0&0&1&\rightarrow&1\\\hline 0&0&2&\rightarrow&2\\\hline 0&1&1&\rightarrow&0\\\hline 0&2&2&\rightarrow&0\\\hline 0&1&2&\rightarrow&0\\\hline 1&2&2&\rightarrow&1\\\hline 1&1&2&\rightarrow&2\\\hline 1&1&1&\rightarrow&1\\\hline 2&2&2&\rightarrow&2\\\hline \end{array} $$The parity of number of $1$-number of $2$ does not change during this operations. Hence there will exist $0,0$ or $1,2$ at the end of the game as desired.$\blacksquare$ What about $1,1$ or $2,2$?

bin_sherlo wrote: We claim that $B$ wins. Rephrased Problem: There are $675$ times $1$,$675$ times $2$ and $674$ times $0$ on a board. On each turn, $A$ chooses $a,b,c$ and $B$ writes one of $(a+b+c)+\{a,b,c\}$ and erases $a,b,c$. After some turns, there will be two numbers on the board. If sum of the numbers written on the board is a multiple of $3$, then $B$ wins and $A$ wins otherwise. Who wins? We work on modulo $3$. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 0&0&0&\rightarrow&0\\\hline 0&0&1&\rightarrow&1\\\hline 0&0&2&\rightarrow&2\\\hline 0&1&1&\rightarrow&0\\\hline 0&2&2&\rightarrow&0\\\hline 0&1&2&\rightarrow&0\\\hline 1&2&2&\rightarrow&1\\\hline 1&1&2&\rightarrow&2\\\hline 1&1&1&\rightarrow&1\\\hline 2&2&2&\rightarrow&2\\\hline \end{array} $$The parity of number of $1$-number of $2$ does not change during this operations. Hence there will exist $0,0$ or $1,2$ at the end of the game as desired.$\blacksquare$ As above points out, this is an incomplete solution, and you need to be more careful to avoid 1, 1 and 2, 2 endings. Interestingly, the problem was proposed with this fakesolve as the solution, although the comitee was able to fix it. Indeed, several students fakesolved the problem this way during the exam.