Three players decide to play a game where the loser doubles the amount of money every other player has. They play three rounds and every player loses once. In the end after 3 rounds each one of them has 40 euros in their pockets.
Tell me the initial money each one of them had!
I experimented with some equations like x + y + z = 120 and 4(x-y-z)=40 and 2(2y-(x-y-z)) = 40 and 4z-2(2y-(x-y--z))-4(x-y-z)=40 (You know doing the process with the variables for each variable)
On
Work backwards: At the end of game 3, the players have $$ 40, 40, 40 $$ Assuming player 3 lost game 3, then the start of game 3 saw: $$ 20, 20, 80 $$ Assuming player 2 lost game 2, then the start of game 2 saw: $$ 10, 70, 40 $$ So player 1 lost game 1, which means the start of game 1 saw: $$ 65, 35, 20 $$
Assume they all started with unknown amounts:
$$x, y, z$$ After round one: $$x - y - z, 2y, 2z$$ After round two: $$2(x - y - z), 3y - x - z, 4z$$ After round three: $$4(x - y - z), 6y - 2x - 2z, 7z - x - y$$ Solving this system of equations, gives: $$x = 65,\ y = 35,\ z = 20$$