MHB Math Game HELP: Solve Hard Task with Invariants

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The discussion revolves around a mathematical game involving a positive whole number that is transformed based on its parity. If the number is even, it is halved; if odd, one can choose between two transformations, \(3n-1\) or \(3n+1\). The connection to the unsolved Collatz conjecture is highlighted, with a specific variation allowing for the choice between the two transformations. A key insight is that selecting the transformation that results in a multiple of 4 leads to a sequence of steps that reduces the number, ultimately suggesting that reaching 1 is possible. The conversation emphasizes the importance of strategic choice in the transformations to ensure a decrease in value.
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HELP - math game

A positive whole number was written on the board. In each step we rub out the number $$n$$ (written on the board) and we write a new one. If number $$n$$ is even, then we write number $$\frac{n}{2}$$ on the board. If number $$n$$ is odd, then we choose one of the numbers: $$3n-1$$ or $$3n+1$$ and we write it down on the board. Decide, if after finite amount of steps, we can obtain the number 1 one the board (no matter which number was written on the board at the beginning).

Please help. I can't figure it out. I know that it can be connected to invariants.
 
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Deveno said:
This is the Collatz conjecture. It is currently unsolved.

Yes, I found it but there is one extra variation. You can choose either $$3n+1$$ or $$3n-1$$. Doesn't it have an influence on the result?
 
If at some stage of the game you have an odd number $n$ then one of the numbers $3n\pm1$ will be a multiple of $4$. Choose that one. Then the next two steps will take you to the number $\dfrac{3n\pm1}4$, which is strictly smaller than $n$. This process of making numbers strictly smaller will inevitably bring you down to $1$ eventually.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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