Not that I was serious about the following question, but I think it is a must-to-ask as a follow-up to [this MO post](https://mathoverflow.net/questions/94525/equal-digit-sums). 


For integer $n\ge0$, let $s(n)$ denote the sum of the digits in the decimal representation of $n$.



> 
> Is it true that for any integer $a,b>0$, the ratio of which is not a power of $10$, the set of all those $n\ge 0$ with $s(an)=s(bn)$ has zero density? 
> 
> 
> 



The digit sum: $s(na)=s(nb)$
OK, by Seva's request I'm getting somewhat more serious :) Fix $a$, $b$. Take large $M$ to be chosen shortly. Take a $3MN$ 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. Let us call these $3M$-digit pieces $n\_k$ so $n=\sum\_{k=0}^{N-1}10^{3Mk}n\_k$. Consider $n\_k$ as independent random variables uniformly distributed in the set $\{0,1,2,\dots,10^{3M}-1\}$. By the law of large numbers, with probability close to $1$, there are about $N\cdot 10^{-2M}$ pieces $n\_k$ that start and end with $M$ zeroes (with the middle $M$ digits being anything). Call the numbers $n$ that satisfy this property typical. The typical numbers can be split into groups according to exactly which $k$ correspond to such pieces (denote by $K$ the set of all such $k$) and what number is formed by the digits outside these 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 typical numbers the portion of the solutions is small.


As Fedor explained, if $M$ exceeds the number of digits in $a$, we have $s(an)=s(an')+\sum\_{k\in K}s(an\_k)$ and similarly for $b$. Thus
$$
s(an)-s(bn)=s(an')-s(bn')+\sum\_{k\in K}X\_k
$$
with 
$$
X\_k=s(an\_k)-s(bn\_k)
$$
The 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 any fixed integer-valued random variable with finitely many values that is not constant, then $\lim\_{Q\to\infty}\sup\_{S\in\mathbb Z} P\{\sum\_{k=1}^Q X\_k+S=0\}=0$ where $X\_k$ are i.i.d. random variables equidistributed with $X$.


Since the cardinality $Q$ of $K$ for every typical group is huge when $N$ is large enough, we conclude that the above displayed equation has very little chance to hold in every 
typical group $G(K,n')$ 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. 


I apologize if this edit rendered some comments meaningless.


