Problem: Positive integers $a$ and $N$ are fixed,
and $N$ positive integers are written on a blackboard.
Alice and Bob play the following game.
On Alice's turn, she must replace some integer $n$ on the board with $n+a$,
and on Bob's turn he must replace some even integer $n$ on the board with $n/2$.
Alice goes first and they alternate turns.
If on his turn Bob has no valid moves, the game ends.

After analyzing the $N$ integers on the board, Bob realizes that,
regardless of what moves Alice makes,
he will be able to force the game to end eventually.
Show that, in fact, for this value of $a$ and these $N$ integers on the board,
the game is guaranteed to end regardless of Alice's or Bob's moves.

Solution: For $N=1$, there is nothing to prove since each player has only one option each turn.
We address $N \ge 2$ only henceforth.
Let $S$ denote the numbers on the board.

\begin{claim*}
  When $N \ge 2$, if $\nu_2(x) < \nu_2(a)$ for all $x \in S$,
  the game must terminate no matter what either player does.
\end{claim*}
\begin{proof}
  The $\nu_2$ of a number is unchanged by Alice's move
  and decreases by one on Bob's move.
  The game ends when every $\nu_2$ is zero.

  Hence, in fact the game will always terminate in exactly
  $\sum_{x \in S} \nu_2(x)$ moves in this case,
  regardless of what either player does.
\end{proof}

\begin{claim*}
  When $N \ge 2$, if there exists a number $x$ on the board such that
  $\nu_2(x) \ge \nu_2(a)$, then Alice can cause the game to go on forever.
\end{claim*}
\begin{proof}
  Denote by $x$ the first entry of the board (its value changes over time).
  Then Alice's strategy is to:
  \begin{itemize}
    \ii Operate on the first entry if $\nu_2(x) = \nu_2(a)$
    (the new entry thus has $\nu_2(x+a) > \nu_2(a)$);
    \ii Operate on any other entry besides the first one, otherwise.
  \end{itemize}
  A double induction then shows that
  \begin{itemize}
    \ii Just before each of Bob's turns, $\nu_2(x) > \nu_2(a)$ always holds; and
    \ii After each of Bob's turns, $\nu_2(x) \ge \nu_2(a)$ always holds.
  \end{itemize}
  In particular Bob will never run out of legal moves,
  since halving $x$ is always legal.
\end{proof}