If 3 divides a number it divides the sum of the digits, why is it that when doing 3*2^n we seem to skip the sums that 9 divides? Here are some $3*2^n$ numbers with their sums:

3           3 

6           6 

12          3 

24          6 

48          12 

96          15 

192         12 

384         15 

768         21 

1536        15 

3072        12 

6144        15 

12288       21 

24576       24 

49152       21 

98304       24 

196608      30 

393216      24 

786432      30 

1572864     33 

3145728     30 

6291456     33 

12582912    30 

25165824    33 

50331648    30 

100663296   33 

201326592   30 

402653184   33 

805306368   39 

1610612736  33 

3221225472  30 

6442450944  42 

12884901888     57 

25769803776     60 

51539607552     48 

103079215104    33 

206158430208    39 

412316860416    42 

824633720832    48 

1649267441664   60


Why are 9, 18, and 27 not included? Doing a python script up to $3*2^{1000}$ gave no sums that were divisible by 9.
Given an integer $n$ let $S(n)$ denote the sum of digits of $n$
$$n=a_k\cdots a_1a_0\implies n=\sum_{i=0}^{k}a_i10^i \text{ and } S(n)=\sum_{i=0}^ka_i$$


  Claim 1:
  $n$ is divisible by $3$ if and only if $s(n)$ is divisible by $3$


proof The following congruences are true modulo $3$ because $10^i\equiv 1\mod 3$ for all $i$

$$3\text{ divides }n \Leftrightarrow n \equiv 0\Leftrightarrow \sum_{i=0}^{k}a_i10^i\equiv 0\Leftrightarrow \sum_{i=0}^{k}a_i1^i\equiv 0 \Leftrightarrow S(n)\equiv 0  \Leftrightarrow 3\text{ divides }s(n)$$


  Claim 2:
  $n$ is divisible by $9$ if and only if $s(n)$ is divisible by $9$


Proof (the same as for claim1)

Now we can conclude using claim $1$  and claim $2$ that $s(3\cdot 2^n)$ is divisible by $3$ and not divisible by $9$
For a number to be divisible by $9$ it needs to have two $3$s in it's prime factorization.

With that in mind it's pretty obvious $3\cdot2^n$ doesnt divide $9$.

Or equivalently 3 doesn't divide $2^n$.

Try and see if you can prove that.