Bruno has a winning strategy Notation: >> $[x, y]$ is the interval of all the natural numbers from $x$ to $y$ inclusive. >> $(x, y)$ is the interval of all the natural numbers from $x$ to $y$ inclusive, and with the same parity than $x$ and $y$. This requires that $x$ and $y$ have the same parity. Let $S_A = (2)\cup(5, 13)\cup(28, 54)\cup(57, 167)\cup(336, 670)$ Let $S_B = (3)\cup(4, 12)\cup[14, 27]\cup(29, 55)\cup(56, 166)\cup[168, 335]\cup(337, 669)\cup(671, 2010)$ Notice that $S_A\cup S_B = [2, 2010]$ and $S_A\cap S_B = \emptyset$ The strategy is to always have a number from $S_A$ written in the board when Ana's turn starts. Bruno can achieve this as follows: If when Bruno's turn starts, the number written in the board is a number that belongs to: >> $(3)$ then he does $*3$ >> $(4, 12)$ then he does $+1$ >> $[14, 27]$ then he does $*2$ >> $(29, 55)$ then he does $*3$ >> $(56, 166)$ then he does $+1$ >> $[168, 335]$ then he does $*2$ >> $(337, 669)$ then he does $+1$ >> $(671, 2010)$ then he does $*3$ and wins Check that by applying any of the three operations to a number in $S_A$ we always get a number in $S_B$