Tom and Burt are to take turns removing either 1 or 2 marbles from a circle. There are 7 marbles. The person selecting the last marble is the loser. If Tom goes first and each of them makes the best possible removals, then who will lose the game?
To determine who will lose the game, we can simulate the turns and analyze the strategy.
We start with 7 marbles. Since Tom goes first, he has the advantage of making the initial move.
1. Tom takes 1 marble. Remaining marbles = 6.
2. Burt takes 2 marbles. Remaining marbles = 4.
3. Tom takes 1 marble. Remaining marbles = 3.
4. Burt takes 2 marbles. Remaining marbles = 1.
5. Tom takes the remaining marble.
Based on this simulation, Tom wins the game.
To solve this kind of problem, we can use the concept of "nim-sum". The nim-sum of the pile of marbles is calculated by finding the XOR (^) of the number of marbles at each turn.
In this case, the nim-sum can be calculated as follows:
nim-sum = 7 ^ 6 ^ 3 ^ 1 = 1
If the nim-sum is equal to 0, then the person who starts the game will lose. Otherwise, the person who starts will win.
Since the nim-sum of 1 is not equal to 0, Tom will win the game.
Therefore, Tom will be the winner in this game if both players make the best possible moves.