Just to be concrete, consider the digits to be binary. Hasse showed that among all the primes, only a fraction of $17/24 < 1$ divide a number of the form $2^n+1$. As a result, the integers that divide a number with just two non-zero digits have zero density.

On the other hand, since $A + A = \mathbb{F}_p$ for a typical set $A \subset \mathbb{F}_p$ of cardinality, say, $> p/\log{p}$ (and much lower than that), and because with probability $1$ there are at least as many powers of $2$ mod $p$, we expect a full density of the primes to divide a number of the form $2^m+2^n+1$. So it would seem legitimate to ask the question (apparently paradoxical?) whether the probability might be positive for a random integer to divide some number composed of only three non-zero digits. Note that $2^k-1$ does not have this property for $k > 3$ (all its multiples have digit sum exceding $ck$), and this family already furnishes an infinite set of pairwise co-prime integers not having the property.

Yet it seems strange that a random integer, with positive probability, would have a multiple with bounded digit sum. Is there a better heuristic for what the answer should be? If the statement is plausible, is it altogether impossible to prove? If it is not, would anything change if I raised the number of non-zero digits to, say, ten? On the opposite extreme, why wouldn't most positive integers $N$ require, for all their multiples, as many as $(1+o(1))\frac{\log{N}}{2\log{2}}$ binary ones?

But perhaps the only safe question that I could ask here is whether an asymptotic can be obtained for the number of positive integers up to $X$ having a multiple of the form $2^m+2^n+1$.