Equal digit sums Let $s(a)$ be the sum of decimal digits of a number $a$. Is it known that for any $a\ne b$ exist $n$ such that $s(na)\ne s(nb)$?
 A: Yes, but if we replace condition $a\ne b$ to $a/b\ne 10^k$ for all integers $k$.
Lemma: $s(9n)=9s(n)$ iff decimal digits of $n$ are only 0's and 1's. Else $s(9n)<9s(n)$.
Call two numbers equivalent, if their ratio is a power of 10 (with integer exponent).
Consider numbers 1,11,111,... Two of them are congruent modulo a, hence their diffrence 11...100..0 equals $ka$ for some natural $k$. Replace pair $(a,b)$ to $(ka,kb)$ and then replace new $a$ (old $ka$) to equivalent number of the form 11...1, totally $m$ ones. Also, divide $b$ to maximal possible power of 10.
Now $s(9a)=9s(a)$, hence the same holds for $b$, hence by lemma $b$ also has only 1's and 0's in its decimal representation and the number of ones equals $m$. In particular, $b>a$ and last digit of $b$ equals 1. Now choose $N$ such that $Nb=11\dots 1$, then $Na < Nb$ and due to $s(Na)=s(Nb)$, $Na$ has a digit different from 0 and 1. Then by lemma $s(9Na)<9s(Na)$, while $s(9Nb)=9s(Nb)$. A contradiction. 
I know this as an old problem by Sergey Berlov, by the way. First time I saw it in 1997 in Sochi on the event for high school students I have participated in:)
