Baltic Way 2019

First we notice that for each odd number $u$, $1\le u<500$, the numbers $u,2u,2^{2}u,\dots ,2^{k_u}u$, where $k_u$ is the largest integer such that $u{2}^{k_u}\le 500$, form a string of length $k_u+1$ such that it is only possible to erase each number in the string if you also erase one of its neighbours. We now divide the numbers on the board into strings and count the lengths of these strings. It is not difficult to see that the numbers of strings of different lengths are

length	1	2	3	4	5	6	7	8	9
number of strings	125	62	32	15	8	4	2	1	1

It is easy to see that a configuration with an even number of strings of each length is a losing position: Player $B$ pairs a string with one of the same length, and whenever $A$ makes a turn on one string, $B$ copy that turn using the paired string. This type of configuration we call a symmetric configuration.

Now if we have a winning configuration and add a symmetric configuration it is still a winning configuration: If $A$ is going to play starting from a winning configuration $W$ and we add a symmetric configuration $S$, then $A$ starts by following her winning strategy on $W$. Every time $B$ replies by changing a string in $W$, $A$ continues using her winning strategy on $W$, and every time $B$ replies by using a string in $S$, $A$ continues by a symmetric move on $S$. It is easy to see that this is a winning strategy.

Now we prove that the above configuration is a winning one, and hence that $A$ has a winning strategy. You can not make a turn on a string of length $1$. Hence we just have to prove that the configuration with three strings of length $4$, $8$ and $9$ is a winning strategy. First $A$ erase the two middle numbers in the string of length $9$, and hence reduces it to two strings of lengths $3$ and $4$, leaving a configuration with two strings of length $3$ and $8$ plus a symmetric configuration. Now player $B$ can either play on the string of length $3$, leaving a string of length $8$, or $B$ can play on the string on length $8$ leaving either: one string of length $6$, one string of length $5$, two strings of length $2$ and $4$, two strings of length $3$. This shows how $A$ replies:

$B$ leaves	$8$	$3+6$	$3+5$	$3+2+4$	$3+3+3$
$A$ leaves	$3+3$	$3+3$	$3+3$	$3+2$	$3+3$

Notice that $A$ can destroy a string of length $3$ or $4$ in one turn. The situation $3+2$ is essentially symmetric since you can only make one move on each string. Hence $A$ leaves a symmetric configuration to $B$ in all the cases, and hence $A$ has a winning strategy.