Alan's 3N Game

We will consider the numbers created by Alan and Bob in base $3$ . If Bob's choices are $a_1,a_2,...,a_k$ , then before Bob's k-th move the number is $1a_1a_2...a_{k-1}0$ and after Bob's k-th move it is $1a_1a_2...a_k$ . Since $3^7>1000$ , the maximal length of the number is at most $7$ . The smallest seven-digit number that can appear in the game is $1111110$ . It equals $3^6+3^5+3^4+3^3+3^2+3^1=\frac{3(3^6-1)}{3-1}=1092>1000$ Therefore, all reachable numbers up to $1000$ have six digits or less in base $3.$ Since they only have 6 digits in base 3, they will be less than $3^6 = 729$ , hence are less than 1000. The first digit is fixed at $1$ , the second through fifth digits can be $1$ or $2$ and the last digit can be $0,$ $1$ or $2$ . This gives $2^4\cdot 3$ possibilities. Similarly, there are $2^3\cdot 3$ five-digit numbers, $2^2\cdot 3$ four-digit numbers, $2\cdot 3$ three-digit numbers, $3$ two-digit numbers and $1$ one-digit number ( $1$ itself). So the grand total is $2^4\cdot 3+ 2^3\cdot 3 + 2^2\cdot 3 + 2\cdot 3 + 3 + 1=94$