If $ a_n$ is increasingly divisible by $2$ and not a multiple of $10$ then the sum of its digits goes to infinity 
Let $(a_n)_{n \geq 0}$ be a sequence of positive integers not divisible by 10 such that the number of factors 2 in $a_n$ tends to inﬁnity for $n \to \infty$. Prove that the sum of the digits of an in the decimal system tends to inﬁnity for $n \to \infty$.


What I did:
Feel free to use any method you like, but it was originately meant to be solved with the 10-adic numbers $\mathbb{Z}_{10}$.
Any element $(x_n)_{n \geq 0} \in \mathbb{Z}_{10}$ can be represented as a "number"
$$
\sum_{n \geq 0} c_n 10^n
\quad \text{ for some } c_n \in \{0, 1, \dots, 9 \}
\quad \text{ for each integer } n \geq 0.   
$$
In this way we can identify $\mathbb{Z}$ with a subset in $\mathbb{Z}_{10}$
that we denote by $D$. 
If we equip $\mathbb{Z}/10^n \mathbb{Z}$ with the discrete topology and 
$\prod_{n \geq 1}\mathbb{Z}/10^n \mathbb{Z}$ with the corresponding product topology, $\mathbb{Z}_{10}$ and $D$ can be equipped with the induced topologies since 
$
D \subseteq \mathbb{Z}_{10} \subseteq \prod_{n \geq 1}\mathbb{Z}/10^n\mathbb{Z}
$.
I showed that $D$ is dense in $\mathbb{Z}_{10}$. If we define
$$
v \ : \ D \ \longrightarrow \ \mathbb{R}
\ : \ a \ \longmapsto \frac{1}{\text{number of factors 2 in } a}  
$$ 
This map is continuous on $D$. By density of $D$ it can be extended in a unique way to $\mathbb{Z}_{10}$.
Hoewever, I have no idea how to prove the given statement.

The problem is that I don't know what the number of decimals actually represents in this framework. Could you help me with that?
The question is due to this syllabus, exercise 1.15.
 A: Let $v_2: \mathbb{Z}_{>0} \to \mathbb{Z}_{\geq 0}$ be the function such that $v_2(a)$ denotes the highest power of $2$ that divides $a$. Let $D_n$ be the set of positive integers not divisible by $5$ such that the sum of the the digits does not exceed n. For example
$$
D_1 = \{1\}
$$
and
$$
D_2 = D_1 \cup \{2,11,101,1001,\dots\}.
$$
We will prove the following lemma.

Lemma: For all $n\geq 1$ the set $\left\{v_2(a):a\in D_n\right\}$ is bounded and since it is a subset of $\mathbb{Z}_{\geq 0}$ we can define $$M_n = \max\left\{v_2(a): a\in D_n \right\}.$$

Proof: We will show this by induction. For $n=1$ we get $D_1=\{1\}$, so the statement is clearly true with $M_1=0$.
Now take an arbitrary $n$ and suppose that the lemma holds for $n-1$. Take some $a\in D_n$ and suppose that it has more than $M_{n-1}+1$ digits. Then we can isolate the leftmost digit $d$ of $a$ and write $$a = d\cdot 10^e + r.$$ For example we would write $6348 = 6 \cdot 10^3 + 348$. We get $e > M_{n-1}$ and $r\in D_{n-1}.$ Now divide $a$ by $2^{v_2(r)}$ to get
$$
\frac{a}{2^{v_2(r)}} = d \cdot 5^e \cdot 2^{e-v_2(r)} + \frac{r}{2^{v_2(r)}}.
$$
By definition $\displaystyle\frac{r}{2^{v_2(r)}}$ is odd and since $e-v_2(r)\geq e - M_{n-1}>0$ we find that $d \cdot 5^e \cdot 2^{e-v_2(r)}$ is even. So $\displaystyle\frac{a}{2^{v_2(r)}}$ is not divisible by $2$. This shows that $v_2(a)=v_2(r)$. This shows that adding zeros to make the number larger eventually can't help in creating factors of two. Concretely this shows that
$$
M_n = \max\left(M_{n-1},\max{\left\{v(a):a\in D_n,\text{ $a$ has no more than $M_{n-1}+1$ digits}\right\}}\right).
$$
This last maximum exists because that set is finite. This proves the lemma. $\Box$
Now we can consider your sequence $(a_n)_{n\geq 0}$ and your continuous function $v$ with $v(a)=1/v_2(a)$. Take $B\in \mathbb{Z}_{>0}$. Since $v(a_n)$ goes to zero there is some $N$ such that for $n>N$ we have $v(a_n)<\frac{1}{M_B}$. This means that for $n>N$ we get $v_2(a_n)>M_B$, so $a_n \not\in D_B$. So the sum of the digits of all $a_n$ will be bigger than $B$ from this point. Since $B$ was chosen arbitrarily this shows that the sum of the digits will go to infinity.
A: Let $a\in\Bbb N$ and $a=\sum_{n=0}^Nc_n10^n$ its decimal representation with $c_0,\dots,c_N\in\{0,\dots,9\}$. Suppose $10\nmid a$ and $2^k\mid a$ for some $k\geq1$. Since $10\nmid a$ we have $c_0\ne0$. Suppose that $k\geq4$, i.e. $2^k\geq10$. If $c_1=c_2=c_3=0$ we would get $10^4\mid a-c_0$ and hence $2^4\mid c_0$ which is not possible as $c_0\in\{1,2,\dots,9\}$. Thus, one of $c_1,c_2,c_3$ is not $0$. Next, suppose that $2^k\geq10^4$. If $c_4=c_5=\cdots=c_{k-1}=0$ we would have $10^k\mid a-c_0-c_110-c_210^2-c_310^3$. But then $2^k\mid c_0+c_110+c_210^2+c_310^3$ which is impossible as $0<c_0+c_110+c_210^2+c_310^3<10^4\leq 2^k$. Hence, one of $c_4,c_5,\dots,c_{k-1}$ is non-zero. Next suppose that $2^{k'}\geq10^{k+1}$ and $2^{k'}\mid a$. We see that then one of $c_{k},\dots,c_{k'-1}$ is non-zero.
Proceeding like that we see that given an integer $n$ there is some $k$ such that if $2^k\mid a$ and $10\nmid a$, then at least $n$ digits of $a$ are non-zero. This clearly implies the claim.
