Two players play a game where they alternatively cross out a number from the numbers written on the board ($1-21$). They stop when two numbers are remaining. If thie sum of these two numbers is divisible by $5$, then the first player wins. Otherwise the second player wins. Who has the winning strategy ?
$$ \begin{align} 1 &&2 &&3 &&4 &&0 \\ 1 &&2 &&3 &&4 &&0 \\ 1 &&2 &&3 &&4 &&0 \\ 1 &&2 &&3 &&4 &&0 \\ 1 \end{align} $$ My reasoning is simple. I have conjectured Player B's best strategy and have shown that A has one smart reply to that. Please tell me if this is correct. Replace each number with its residue $\pmod 5$. I define two residues as complementary of they sum up to $0$ like $2$ and $3$. When a player crosses out a number, player B should cross out cross out a number from the same residue class. If Player A crosses out a number from a number from the first class again. What I mean by this if Player 1 crosses out $1$ in the beginning, then player $2$ also crosses out $1$. If Player $2$ crosses out $4$ at some stage in the future, Player B once again crosses out a number of residue $1$. The idea is that Player B only has to focus on three residues. So, Player A can delay B by at most nine moves.After that there won't be three numbers moves. Then, player two wins unless there are two numbers remaking that leave a residue of $0$ with 5. In all other cases, Player two wins.
Here is Player $A$ winning strategy. Strike out one first, and the keep striking out $4$. Then strike out 2 and 3 alternatively. (Opposite of whatever B) strikes). At the end there will be two numbers divisible by 5 and some other number, which A will strike out to win. So, I guess Player $A$ has a winning strategy. Is my reasoning correct and is this the best possible strategy for Player 2.
Player $A$ strikes out $21$ (so that the sum of all the remaining numbers is divisible by $5$), and then uses the following pairing:
$(1,19),(2,18),(3,17),(4,16)$
$(6,14),(7,13),(8,12),(9,11)$
$(5,20),(15,10)$
Whatever $B$ strikes out, $A$ strikes out the other number of the pair, leaving a sum divisible by $5$.