Two friends are playing a game where they break of pieces from a rectangular 10 x 15 candy bar. The game continues until either player gets a 1 x 1 piece. In the first move player A breaks the bar along a line. The second move is that player B chooses one of the two pieces and breaks it along a line (creating 3 total pieces). Third move is that player A can choose any of the three pieces (not just the latest two), and breaks it along a line. The game continues until either player breaks a piece leaving a 1 x 1 piece.
The question is if either of the players have a winning strategy if i) the loser is the player that creates a 1 x 1 piece ii) the winner is the player that creates a 1 x 1 piece
My reasoning is that after move $n$ there will be $n + 1$ pieces. Assuming optimal play there will be only 1 x 2 pieces left when either player is forced into losing in (i). That's exactly 75 pieces, so that's after move 74 which player B makes. That means that player A loses?
I don't really have a good idea of coming up with a strategy for (ii) though, any help of clue would be very much helpful!
For both variants, $A$ has the winning strategy:
$A$'s first move will always be to break the bar in half, into two $5 \times 15$ pieces. Then $A$ will mirror $B$'s moves, so that at the end of $A$'s turn, there will always be an even number of pieces of size $m \times n,$ for every applicable $m,n,$ until $B$ makes a losing move.
For variant $(i),$ we define a losing move to be when $B$ makes a $1 \times 1$ piece. Obviously, at this point, $B$ will have lost with no more input from $A$.
For variant $(ii),$ we define a losing move to be when $B$ makes a $n \times 1$ (or $1\times n$) piece. It's easy to argue that the first time this occurs will happen on $B$'s turn and also that $n > 1$. Therefore, $A$ can break off the end and be the first to create a $1 \times 1$ piece, winning the game.