João and Maria won 3 chocolate bars of $5 \times 3$ divided into $1 \times 1$ squares. So they decide to play a game. João takes one of the bars and divides it into two smaller rectangular bars by cutting it along one of the marked dividing lines between the squares of the bar. Then Maria takes any of the bars and also divides it using one of the marked dividing lines on it. They continue dividing the bars alternately, and the winner is the one who, after their turn, leaves only $1 \times 1$ squares as pieces. Who wins the game?
![](https://cdn.mathpix.com/cropped/2024_05_01_80f7f1185898b39e7876g-14.jpg?height=356&width=796&top_left_y=2338&top_left_x=525)

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Solution

Note that the number of pieces on the table increases by one unit after each move. Thus, after João's first move, there will be 4 pieces on the table, and after Maria's first move, there will be 5 pieces. After João's second move, there will be 6 pieces, and so on. Notice that João always leaves an even number of pieces, and Maria always leaves an odd number. Since there are 3 bars of $5 \times 3$, at the end of the game, there will be $3 \times 3 \times 5=45$ $1 \times 1$ squares left. Since 45 is odd, Maria will win the game.