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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.

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Not that I'm serious about this answer, but yes. Take an $N$ digit number $n$ and split its decimal representation into pieces of length $3M$ where $M$ is large but fixed. A noticeable portion of these pieces will be of the kind $M$ zeroes, something, $M$ zeroes. $M$ zeroes will control the transfers if $M$ is large enough and, conditioning upon the location of these pieces and the rest of the representation, we see that these pieces will give us long sums of independent variables. Now, we need 3 statements:

1) If $X$ is an integer-valued random variable with finitely many values that is not constant, then $\lim_{N\to\infty}\sup_{k\in\mathbb Z} P\{\sum_{i=1}^N X_i=k\}=0$ where $X_i$ are i.i.d. random variables equidistributed with $X$.

2) If $s(an)-s(bn)=C$ for all $n$ (i.e., if we cannot choose $M$ to use Claim 1), then $C=0$.

3) 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 (if I find one while I'm driving to work, I'll post it, but, most likely, someone else will beat me to that).