For a natural number $n$, let $P_n$ the set of sums of each subset of digits in decimal notation of $n$. A number is magic if for each $s \in P_n$, we have $s \ | \ n$. Let's consider a number $\ell$, not divisible by $10$. Then, only a finite number of natural powers of $\ell$ are magic.
My approach: Proof by contradiction principle – logic formalism principle:
$$(\neg q \to \neg p) \leftrightarrow (p \to q)$$
In an infinite set $I$ of natural powers, we have, by combinatorics principles (I think Wan der Waerden?), there is an arbitrary long arithmetic progression in the exponents. So, therefore, considering an "almost infinite" progression (enough long), I'm thinking to find here a contradiction, as arithmetic progressions at the exponents do mean geometric progressions, therefore constant ratios.
I've read also some papers (statistical approach), that are trying to find the lower bound for number of specific digits (i. e. number of $1$s) in powers. And, therefore, try to find divisors that are mandatory in $P_{a^k}$, where $k$ is large enoungh, and here we got a contradiction, as the only divisors of $a_k$ are the powers of divisors of $a$.
However, statistical approach is not a rigurous mathematical proof, at least for my purpose, so, those assumptions could be no more than conjectures about the problem.
Please give me a hint! Any help would be appreciated.
Update: As @Vepir mentioned, "magic" numbers are super Nieven numbers.
Best Answer
$\textbf{Theorem:}$ If $l$ is not a power of $10$ then we have
$$\lim_{n\rightarrow \infty}\text{Digitsum}(l^n) = \infty$$
here $\text{Digitsum}(.)$ of a number is the sum of its respective digits. More details for this theorem can be found in; https://mathoverflow.net/questions/38971/sums-of-digits-of-powers-of-integers.
Now for a contradiction suppose that there are an infinite sequence of positive integers $n_{1}< n_{2},...$ so that $l^{n_{j}}$ is magic. Without loss of generality, due to the above theorem we may assume that $\text{Digitsum}(l^{n_{j}}) \geq j.$ Set
$$A_{j} = \text{ Set of sums of each subset of digits in decimal notation of } l^{n_{j}}$$
$\textbf{Claim 1}:$ $A_{j}$ consists of atleast $\frac{j}{9}$ numbers in the interval $[1, j].$
$\textbf{Claim 2}:$ If some prime $p$ divides an element in $A_{j}$ then it divides $l$.
We see this result when one sees that each element of $A_{j}$ divides $l^{n_{j}}$.
$\textbf{Claim 3}:$ For each $m \in \mathbb{N}$, $\bigcup_{j = 1}^{\infty} A_{j}$ has atleast $\frac{m}{9}$ integers in interval $[1, m]$
$\textbf{Claim 4}:$ For every $v \in \mathbb{N}, \exists p$ prime so that $p > v$ and $p$ divides some element of $\bigcup_{j = 1}^{\infty} A_{j}$.
Suppose for a contradiction that this claim is false. We can choose primes $p_1,p_{2},...,p_{k}$ so that every element of $\bigcup_{j = 1}^{\infty} A_{j}$ is of the form $p_{1}^{\alpha_{1}}...p_{k}^{\alpha_{k}}$. Hence
$$\sum_{x \in \bigcup_{j = 1}^{\infty} A_{j}} \frac{1}{x} \leq \sum_{\alpha_{1}=0}^{\infty}...\sum_{\alpha_{k}=0}^{\infty}\frac{1}{p_{1}^{\alpha_{1}}...p_{k}^{\alpha_{k}}} = \prod_{a=1}^{k}(\frac{p_{a}}{p_{a}-1}) < \infty$$ but by claim 3 we have $$\sum_{x \in \bigcup_{j = 1}^{\infty} A_{j}} \frac{1}{x} \geq \sum_{j=1}^{\infty}\frac{1}{9j} = \infty$$ a contradiction.
Hence by Claim 4 and Claim 2 there are primes that are arbitrarily large that divide $l$ but this is impossible.