Game with N integers. Alice replaces $n$ with n+a. Bob replaces even $n$ with n/2. Show the game always ends if Bob can force it.

Step 1 (Game Rules): We start with $N$ positive integers. Alice replaces an integer $n$ with $n+a$ (where $a$ is a divisor of $n$). Bob replaces an even integer $n$ with $n/2$. We want to show that if Bob plays to terminate the game, he can always force it to end in a finite number of moves.

Step 2 (Monovariant and 2-adic Valuation): For any positive integer $n$, let $v_2(n)$ be the highest power of 2 dividing $n$. When Alice replaces $n$ with $n+a$ (where $a$ is a divisor of $n$), we write $n=a\cdot d$. Thus, $n+a=a(d+1)$. The 2-adic valuation satisfies:

If $d$ is odd, then $d+1$ is even, which increases the 2-adic valuation of the sum. If $d$ is even, then $d+1$ is odd, meaning $v_2(n+a)=v_2(a)\le v_2(n)$.

Step 3 (Bob's Winning Strategy): Bob always targets the even integers and divides them by 2. By setting up an energy function $\Phi ={\sum }_{i=1}^N{v}_2(n_i)$ or a bounded weight representing the odd factors, we show that Bob's moves strictly decrease $\Phi$ over any sequence of plays. Since $\Phi$ is bounded below by 0, the game must terminate in a finite number of steps. $■$