Fun Number Theory Game

It is easier to reverse the rules and look for a path from $100$ to $1$ . When going backward a simple heuristic suffices:

Our two possible moves are $n \rightarrow n/3$ and $n \rightarrow n+1$

From the number $n$

a) if $3 | n$ , move to $n/3$

b) if $3 \nmid n$ , move to $n+ 1$ .

Clearly if $3 \nmid n$ we cannot move to $n/3$ as every number in the forward path is an integer.. So to show the soundness of this approach, it suffices to show that, when possible, it's preferable to divide by $3$ than to add $1$ .
This is fairly obvious: the path of

$n \rightarrow n/3 \rightarrow n/3 +1$

gets us to the same place as

$n \rightarrow n+1 \rightarrow n+2 \rightarrow n+3 \rightarrow n/3+1$ ,

and it takes 2 fewer steps.

So, our backwards path goes $100 \rightarrow 101 \rightarrow 102 \rightarrow 34 \rightarrow 35 \rightarrow 36 \rightarrow 12 \rightarrow 4 \rightarrow 5 \rightarrow 6 \rightarrow 2 \rightarrow 3 \rightarrow 1$

Our optimal path has 12 steps.

we can do the reversed moving from 100 to 1:

$((((100 + 1 + 1) \div 3 + 1 + 1) \div 3 \div 3 + 1 + 1) \div 3 + 1) \div 3 = 1$