I will assume that the reader knows the Collatz ($3n+1$) conjecture.
Terminology: let's say that a natural number $ n $ is a descendant of $ m $ if the Collatz procedure starting at $ m $ eventually leads to $ n $. For example, $ 5 $ is a descendant of $ 7 $ since the Collatz procedure starting at $ 7 $ yields $$ 7 \rightarrow 22 \rightarrow 11 \rightarrow 34 \rightarrow 17 \rightarrow 52 \rightarrow 26\rightarrow 13 \rightarrow 40 \rightarrow 20 \rightarrow 10 \rightarrow 5 $$ Let's also say that $ m $ is an ancestor of $ n $. (So $ 7 $ is an ancestor of $ 5 $.)
Question 1: Is it true that all natural numbers $ n $ have an ancestor that is a multiple of $ 3 $?
Question 2: If Question 1 is non-trivial, does anyone happen to know if it implies the Collatz conjecture? On the other hand, if it is trivial, or at least proven, can they point me to proof?
Question 3: Assuming the answer to Question 1 is affirmative, can such an ancestor be found by repeatedly applying the "greedy" reverse-collatz function $$ g(n) = \begin{cases} \frac{n-1}{3} & n \cong 4\ (\mathrm{mod}\ 6) \\ 2n & n \cong 1, 2,\mathrm{or}\ 5\ (\mathrm{mod}\ 6) \end{cases} $$
I find it interesting to note that, as wonderfully rich as is the topology of the collatz "tree" (whose topology is described by the ancestor/descendant relationship), the topology of the ancestor tree is trivial above any number which is a multiple of 3. (The tree does not branch above multiples of 3.) So an affirmative answer to Question 1 puts some interesting restrictions on the topology of this grand tree.
For positive integer $\ m\ $ , we need a positive integer $\ n>m\ $ with $\ 3\mid n\ $, such that the collatz-sequence beginning with $\ n\ $ contains $\ m\ $.
So, question $1$ can be answered with "yes".
Not sure about question $3$