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 $s(na)\ne s(nb)$?
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asked Apr 19, 2012 at 13:15
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2$\begingroup$ Maybe the following properties of the function $s$ could be useful; $s(a+b) \le s(a) + s(b)$ and $s(a)s(b) \le s(a)s(b)$. $\endgroup$– WoettCommented Apr 19, 2012 at 13:25
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1$\begingroup$ And of course, the sum of digits of $a$ is divisible by $3$ (or $9$) if and only if $a$ itself is divisible by $3$ (or $9$). So $n = 1$ suffices if exactly one of $a$ and $b$ is a multiple of $3$. $\endgroup$– WoettCommented Apr 19, 2012 at 13:30
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9$\begingroup$ Counterexample: $a=1$, $b=10$. But it's probably true (and may be reasonably easy) that the only counterexamples are those for which $b/a$ is a power of $10$. $\endgroup$– Noam D. ElkiesCommented Apr 19, 2012 at 13:40
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5$\begingroup$ If the greater of $a$ and $b$ is coprime to $10$ then there exists $n$ such that $na = 10^c - 1$ for some $c$, and then clearly $s(na)>s(nb)$. [This argument would already prove the guess if the base were prime.] It may be possible to expand on this trick to deal with even numbers and multiples of $5$. $\endgroup$– Noam D. ElkiesCommented Apr 19, 2012 at 14:51
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2$\begingroup$ For what it's worth, one can assume that at least one of $a,b$ is coprime to 10: if $a$ and $b$ are related as in the question, and $a$ is even, then $5a$ and $5b$ are also related as in the question, and hence so are $a/2$ and $5b$. Arguing like this one can reduce to the case where one of $a,b$ is coprime to 10 and the other is not a multiple of 10, and then one has to prove $a=b$. $\endgroup$– Kevin BuzzardCommented Apr 19, 2012 at 18:58
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1 Answer
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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:)
answered Apr 19, 2012 at 20:20
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$\begingroup$ I don't quite follow the "hence $b$ also has only 1's and 0's" conclusion. What about $a=11$ and $b=2$? $\endgroup$ Commented Apr 19, 2012 at 20:34
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$\begingroup$ we must have $s(9b)=9s(b)$, for $b=2$ it is not so. $\endgroup$ Commented Apr 19, 2012 at 20:39
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2$\begingroup$ You can shorten your proof by Elkies' trick: choose $N$ such that $Nb=99\dots9$, then $Nb>Na$, so $s(Nb)>s(Na)$. A contradiction. $\endgroup$ Commented Apr 19, 2012 at 20:40
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$\begingroup$ Oh sorry! My stupid mistake! $\endgroup$ Commented Apr 19, 2012 at 20:41
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$\begingroup$ Perhaps it would be useful to emphasize that $s(n+m)\leq s(n)+s(m)$ with equality iff there is no digit carry when adding $n$ and $m$, cf. Barry's comment and Woett's very first comment. $\endgroup$ Commented Apr 19, 2012 at 20:42