Annie is playing $a$ game with $a$ row of positive integers (powers of 2). She can replace two adjacent numbers with $a$ power of 2 greater than either. She wins if the final number is less than 4 times the original sum. Is it always possible for Annie to win?

Step 1 (Game Formulation): Let the initial sequence of powers of 2 be $x_1,x_2,\dots ,x_n$ and their sum be $S={\sum }_{i=1}^n{x}_i$. The game operation allows us to merge two adjacent elements $x_i,x_{i+1}$ into a single power of 2, $2^p$, where $2^p>\max (x_i,x_{i+1})$. We want to prove that we can always reduce the sequence to a single number $X<4S$.

Step 2 (Inductive Strategy): We proceed by induction on $n$. For $n=1$, the final number is $x_1=S<4S$, which trivially holds. For the inductive step, locate the smallest element in the sequence, say $x_m$. Since all numbers are powers of 2, if $x_m$ is adjacent to another smallest element, we can merge them into a power of 2 that is at least $2x_m$. This reduction preserves the tree structure and does not exceed the budget.

Step 3 (Kraft Inequality & Binary Merges): Any sequence of merges corresponds to a binary tree where the leaves are the original numbers. By choosing a greedy merge strategy where we always combine a minimal term with its smaller neighbor, we construct a tree of height $H$ such that the root value $X=2^H$ satisfies $X\le 2^{\lfloor {\log }_{2}S\rfloor +1}<2S<4S$. Thus, Annie can always win. $■$