National Olympiad of Argentina

The first player $A$ has a winning strategy.

One may reduce $1,2,\dots ,2010$ modulo $3$, so assume that the initial numbers are $1,2,0,\dots ,1,2,0$ (670 triples $1,2,0$).

Let $A$ divide the $2009$ gaps between numbers into one group of $2$ (a double), corresponding to the initial $1,2,0$, and $669$ groups of $3$ (triples), each corresponding to $4$ consecutive $0,1,2,0$. $A$ makes his first move in a triple. Then he can ensure to be the one making the last move in each group as follows:

a) If $B$ moves in the double, $A$ repeats the same move with the second gap in the double.

b) If $B$ is the one to move first in a triple, $A$ moves first in another triple.

c) If $B$ is the one to move second in a triple, $A$ makes the third remaining move in the same triple.

Observe that $A$ can always stick to these rules because there are an odd number of triples ($669$) and his initial move is in one of them. Hence if several moves were made according to a)–c) and it is $B$'s turn to play, there are an even number of triples where no move was made; and $1$ or $3$ moves were made in each remaining triple.

In case b) $A$'s move can be arbitrary. In case c), where $A$ completes a triple $T$, he plays to ensure the following. If eventually $T$ has the form $0\times 1+2\times 0$ then the two $\ast$ are the same; and if $T$ has the form $0\times 1\times 2\times 0$ then one of the two $\ast$ is $\times$. This is always possible by direct verification.

After all moves are complete we obtain the sum $S$ of several products, some of them possibly with one factor. If a product in $S$ does not contain $0$ then it can be $1$, $2$ or $1\times 2$; however $1\times 2$ is excluded. Indeed, such a $1\times 2$ cannot come from a triple $0\times 1\times 2\times 0$ because by the previous paragraph one of the two $\ast$ is $\times$. Similarly if a $1\times 2$ comes from the double $1\times 2\times 0$ then $\ast$ is $\times$.

So a product in $S$ without a $0$ is either a $1$ or a $2$, meaning that it is contained in a part $...+1+\mathrm{2...}$ or $...1+2+...$ of $S$. If the $1$ and the $2$ are in a triple $0\times 1+2\times 0$ then the triple is $0+1+2+0$ as the two $\ast$ are the same, and one is a $+$. If the $1$ and the $2$ are in the double $1+2+0$ then the double is $1+2+0$ (by a)). We conclude that the products without a $0$ in $S$ come in pairs of the form $...+1+2+...$ or $1+2+...$ (for the double). It follows that $S$ is divisible by $3$, so $A$ wins.