It is clear that $1$ remains at the end of the procedure. Let the largest power of two that occurs in the table be $2^k$. (There is at least one power of two, since $x_1 = 1$, and so $2$ appears in the first row of the table.) We note that we must have that $2^k = 2x_m$ for some $m$. This is because we can not have $2^k = 3x_m$, and if $2^k = x_m$, then $2x_m$ is a larger power of two that appears in the table. We see that this implies that $2^k$ appears exactly once in the table, and so remains at the end of the procedure. (There is a unique value for $x_m$ such that $2^k = 2x_m$.) Similarly, the largest power of $3$ in the table remains at the end of the procedure. Consider the sequence $x_1, x_2, \dots, x_9$ given by $x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4, x_5 = 8, x_6 = 9, x_7 = 12, x_8 = 18, x_9 = 27$. By drawing the table and crossing off duplicate entries, we see that only the entries $1$, $16$, and $81$ remain. Now suppose that for some $n$ and some choice $x_1, x_2, x_3, \dots, x_n$, we have that only $1$, $2^a$ and $3^b$ remain at the end of the procedure. Consider the sequence $y_1, y_2, \dots, y_{3n}$ defined as follows: $y_i = x_i$ if $i \leq n$, $y_i = 2^a x_{i - n}$ if $n < i \leq 2n$ and $y_i = 3^b x_{i - 2n}$ if $2n < i \leq 3n$. When applying the procedure to $y_1, y_2, \dots, y_n$, the numbers $1, 2^a, 3^b$ remain. When applying the procedure to $y_{n + 1}, y_{n + 2}, \dots, y_{2n}$, the numbers $2^a, 2^{2a}, 2^a \cdot 3^b$ remain. When applying the procedure to $y_{2n + 1}, y_{2n + 2}, \dots, y_{3n}$, the numbers $3^b, 2^a \cdot 3^b, 3^{2b}$ remain. Thus when applying the procedure to $y_1, y_2, \dots, y_{3n}$, the numbers $1, 2^{2a}, 3^{2b}$ remain. Edit: for part b), we could just use $n = x_1 = 1$ as the base case