[Bulgaria 2005]
For positive integers t, a, b, a (t, a, b)-game is a two player game defined by the following rules. Initially, the number t is written on a blackboard. In his first move, the first player replaces t with either t – a or t – b. Then, the second player subtracts either a or b from this number, and writes the result on the blackboard, erasing the old number. After this, the first player once again erases either a or b from the number written on the blackboard, and so on. The player who first reaches a negative number loses the game. Prove that there exist infinitely many values of t for which the first player has a winning strategy for all pairs (a, b) with (a + b) = 2005.
Note that the players subtract a or b from the given number so if they were to alternate the number would keep reducing by 2005 each time. Its easy for any player to know whether the number substracted was a or b because the previous number and the new number are both on the board one after the other. The winning player has to ensure that the number reduces by 2005 by alternating a or b as per the previous move of the opponent.
In other words, if the winning player has a strategy for x then he has a winning strategy for x + 2005k
Since k can be a multiple of 2005, there are infinitely many numbers possible for this player with the above winning strategy.
The players reduce the numbers by a or b so ultimately there is a value v1 or v2 possible before the number becomes negative. Every other number on the board is therefore a multiple of v1 + a , v1 + b, v1 + a+ b or v2 +a , v2 + b and v2 +a + b.
If it is v2, we call it a favorable position and the first player wins for v2. we say that the strategy works for him.
If it is v1 then the first player. he will win only if v1 +a + b = v2 and v1+a and v1+b are not favorable but that means v1 + a - b and v1 + b - a are favorable but that means v1 + a and v1 + b are favorable which is a contradiction.
For positive integers t, a, b, a (t, a, b)-game is a two player game defined by the following rules. Initially, the number t is written on a blackboard. In his first move, the first player replaces t with either t – a or t – b. Then, the second player subtracts either a or b from this number, and writes the result on the blackboard, erasing the old number. After this, the first player once again erases either a or b from the number written on the blackboard, and so on. The player who first reaches a negative number loses the game. Prove that there exist infinitely many values of t for which the first player has a winning strategy for all pairs (a, b) with (a + b) = 2005.
Note that the players subtract a or b from the given number so if they were to alternate the number would keep reducing by 2005 each time. Its easy for any player to know whether the number substracted was a or b because the previous number and the new number are both on the board one after the other. The winning player has to ensure that the number reduces by 2005 by alternating a or b as per the previous move of the opponent.
In other words, if the winning player has a strategy for x then he has a winning strategy for x + 2005k
Since k can be a multiple of 2005, there are infinitely many numbers possible for this player with the above winning strategy.
The players reduce the numbers by a or b so ultimately there is a value v1 or v2 possible before the number becomes negative. Every other number on the board is therefore a multiple of v1 + a , v1 + b, v1 + a+ b or v2 +a , v2 + b and v2 +a + b.
If it is v2, we call it a favorable position and the first player wins for v2. we say that the strategy works for him.
If it is v1 then the first player. he will win only if v1 +a + b = v2 and v1+a and v1+b are not favorable but that means v1 + a - b and v1 + b - a are favorable but that means v1 + a and v1 + b are favorable which is a contradiction.
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