MHB Can Kolya Win at the Stone Pile Game?

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Kolya and Vitya are playing a game with 31 stones where players take turns dividing piles of stones until only single stones remain. The objective is to leave all piles with one stone after a player's turn. The discussion revolves around whether Kolya can guarantee a win regardless of Vitya's moves. Participants are analyzing the game's mathematical strategy by considering smaller numbers of stones to understand winning and losing positions. Ultimately, the consensus leans towards Kolya not being able to win every time due to the inherent complexities of the game.
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KOLYA AND VITYA PLAY THE FOLLOWING GAME. THERE IS A PILE OF 31 STONES ON THE TABLE. THE BOYS TAKE TURNS MAKING MOVES AND KOLYA BEGINS. IN ONE TURN A PLAYER DIVIDES EVERY PILE WHICH HAS MORE THAN ONE STONE INTO TWO LESSER ONES. THE PLAYER WHO AFTER HIS TURN LEAVES ALL PILES WITH ONLY ONE STONE IN EACH WINS. CAN KOLYA WIN NO MATTER HOW VITYA PLAYS?

I'm pretty sure that Kolya can't win every time but I'm a bit confused on the actual math behind it, would love some explanations. Thanks!
 
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zen said:
KOLYA AND VITYA PLAY THE FOLLOWING GAME. THERE IS A PILE OF 31 STONES ON THE TABLE. THE BOYS TAKE TURNS MAKING MOVES AND KOLYA BEGINS. IN ONE TURN A PLAYER DIVIDES EVERY PILE WHICH HAS MORE THAN ONE STONE INTO TWO LESSER ONES. THE PLAYER WHO AFTER HIS TURN LEAVES ALL PILES WITH ONLY ONE STONE IN EACH WINS. CAN KOLYA WIN NO MATTER HOW VITYA PLAYS?

I'm pretty sure that Kolya can't win every time but I'm a bit confused on the actual math behind it, would love some explanations. Thanks!

Hi zen! Welcome to MHB! (Smile)

Let's work back from the end.
Suppose there is only 1 stone, who will win?
What with 2 stones? And 3?

Now it becomes more interesting.
Suppose we have 4 stones, to win, we need a move that will bring us to a state that is guaranteed to lose.
Is that possible? (Wondering)
 
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