Argentine National Olympiad 2015

Player $B$ wins in part a); player $A$ wins in part b). Note that, by the rules, the player to move can always add $1$ to one of the current numbers.

For $a=6$, $b=7$ let $A$ start the game by adding $1$ to $6$, leading to the pair $7,7$. If $a=3$, $b=5$, $A$'s first move is forced to be adding $1$ to one of the numbers because they are both primes. If $A$ adds it to $3$ then $B$ adds $1$ to the obtained $4$ and obtains $5,5$. And if $A$ adds $1$ to $5$ the new pair is $3,6$. Since $3$ is a proper divisor of $6$, $B$ can add it to $3$ and obtain $6,6$. Hence in part b) $A$ can obtain two equal numbers with his first move; $B$ can achieve the same with his first move in part a). Note that no one can win in one move in positions $a,a$ where $a=5,6,7$.

Now it suffices to show that any position with two equal numbers $a,a^{\prime }$, where $a\le \frac{2}{3}\cdot 2014$, is losing, i.e., the player $X$ to move in this position loses. Hence $X$ obtains a new number $a^{\prime }=a+d>a$ where $d|a$ and $d\le \frac{a}{2}$, so $a^{\prime }\le \frac{3a}{2}\le 2014$.

Hence $X$ does not win on his first move in the position $a,a^{\prime }$. Now the strategy of the opponent $Y$ is as follows. If one of $a$ and $a^{\prime }$ has a proper divisor $d^{\prime }$ such that $a^{\prime }+d^{\prime }\ge 2015$, $Y$ adds $d^{\prime }$ to $a^{\prime }$ and wins. If not, $Y$ repeats the previous move of $X$, adding $d$ to $a$; note that this is possible. This yields a new pair $a^{\prime },a^{\prime }$, of equal numbers, with $a^{\prime }>a$. Observe that now $X$ cannot reach $2015$ in one move, otherwise $Y$ would have done so on his previous move. Hence $X$ has to obtain again a pair of different numbers $a^{\prime },a^{\prime \prime }$, with $a^{\prime }<a^{\prime \prime }\le 2014$. Then $Y$ repeats the same: he either reaches $2015$ in one move if this is possible, or obtains the pair $a^{\prime \prime },a^{\prime \prime }$ from which $X$ cannot win in one move. If $Y$ follows this strategy, the numbers $a,a^{\prime },a^{\prime \prime },\dots$ in the process increase, and $X$ never makes a winning move. There will come a moment when a proper divisor $d$ of $a$ in the current pair $a,a$ is such that $a+d\ge 2015$; then $Y$ wins on that move.