Not that
OK, by Seva's request I'm getting somewhat more serious about this answer, but yes:) Fix $a$, $b$. Take large $M$ to be chosen shortly. Take an a $N$ 3MN$ digit number $n$ (something from $0$ to $10^{3MN}-1$, written with head zeroes if necessary) and split its decimal representation into pieces of length $3M$ where $M$ is large but fixed. A noticeable portion of Let us call these $3M$-digit pieces will be of $n_k$ so $n=\sum_{k=0}^{N-1}10^{3Mk}n_k$. Consider $n_k$ as independent random variables uniformly distributed in the kind set $M$ zeroes, something\{0,1,2,\dots,10^{3M}-1\}$. By the law of large numbers, with probability close to $M$ zeroes. 1$, there are about $N\cdot 10^{-2M}$ pieces $n_k$ that start and end with $M$ zeroes will control (with the transfers if middle $M$ is large enough and, conditioning upon digits being anything). Call the location of these numbers $n$ that satisfy this property typical. The typical numbers can be split into groups according to exactly which $k$ correspond to such pieces and (denote by $K$ the rest set of all such $k$) and what number is formed by the representation, we see that digits outside these pieces will give us long sums groups (that number is $n'=\sum_{k\notin K}10^{3Mk}n_k$). We need to show that in each group $G=G(K,n')$ of independent variablestypical numbers the portion of the solutions is small.Now
As Fedor explained, if $M$ exceeds the number of digits in $a$, we need 3 statements:
1have $s(an)=s(an')+\sum_{k\in K}s(an_k)$ and similarly for $b$. ThusThe random variables $X_k$ are i.i.d. and their distribution is completely determined by $a,b$, and $M$.
Suppose that the equality $s(an)=s(bn)$ fails for at least one $n\ge 0$. Then (since it holds for $n=0$) $X_k$ is not a constant for sufficiently large $M$ and we can use the following
Probabilistic Claim. If $X$ is an any fixed integer-valued random variable with finitely many values that is not constant, then $\lim_{N\to\infty}\sup_{k\in\mathbb \lim_{Q\to\infty}\sup_{S\in\mathbb Z} P\{\sum_{i=1}^N X_i=k\}=0$ P\{\sum_{k=1}^Q X_k+S=0\}=0$ where $X_i$ X_k$ are i.i.d. random variables equidistributed with $X$.
2) If
Since the cardinality $s(an)-s(bn)=C$ Q$ of $K$ for all every typical group is huge when $n$ (i.e., if N$ is large enough, we cannot choose $M$ conclude that the above displayed equation has very little chance to use Claim 1), then hold in every typical group $C=0$.
3) If G(K,n')$ and, thereby, overall.
However, if $s(an)=s(bn)$ for all $n$, then $a$ and $b$ differ only by the number of zeroes in the end.
Claim 3 is the only non-obvious one.
I do not have a decent proof of it yet (apologize if I find one while I'm driving to work, I'll post it, but, most likely, someone else will beat me to that)this edit rendered some comments meaningless.
