Timeline for Equal digit sums
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2012 at 20:00 | comment | added | GH from MO | I like this version better. Still, you can simplify it by Elkies' trick: $9Na<9Nb$ implies directly that $s(9Na)<s(9Nb)$, no need to invoke the Lemma again. | |
| Apr 20, 2012 at 16:30 | comment | added | fedja | Ah, all right. Sometimes I get totally confused with simple but not explicitly mentioned steps :). OK, agreeing and voting up then... | |
| Apr 20, 2012 at 15:15 | comment | added | Fedor Petrov | @all: I tried to make the exposition complete and self-contained. | |
| Apr 20, 2012 at 15:13 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
some details added
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| Apr 20, 2012 at 14:17 | comment | added | GH from MO | @fedja: I continue. Now $s(a)=s(b)$ gives $b>a$, so the argument can be finished by Elkies' trick. We choose $N$ such that $Nb=10^d-1$, then $Nb>Na$, so $s(Nb)>s(Na)$, a contradiction. | |
| Apr 20, 2012 at 14:12 | comment | added | GH from MO | @fedja: Fedor Petrov's proof is OK, there is no casework to do. First of all, as Kevin Buzzard pointed out in a comment to the original post, we can assume that $a$ is coprime to $10$. Then $n\cdot 9a=10^c-1$ for some integer $n>0$, so replacing $a$ by $na$ (resp. $b$ by $nb$) brings us to the situation that Fedor Petrov starts with. Then $s(9a)=9s(a)$, hence $s(9b)=9s(b)$, which means that there is no carry when multiplying $b$ by $9$ in the decimal system. So $b$ only has digits $0$ and $1$. The digits $0$ at the end can be deleted, so $b$ ends with $1$, hence it is coprime to $10$. | |
| Apr 20, 2012 at 13:38 | comment | added | fedja | That all gives only that $a=2^kC$, $b=5^mC$. This much can be achieved in many ways. The devilish casework is after that and I don't know a decent proof (not that I thought of it too hard though) | |
| Apr 20, 2012 at 10:48 | comment | added | GH from MO | @Alexey: There is no gap, just more details should be added. $s(9a)=9s(a)$ implies $s(9b)=9s(b)$, which means that there is no carry when multiplying $b$ by $9$ in the decimal system. So $b$ only has digits $0$ and $1$. The digits $0$ at the end can be deleted, so $b$ ends with $1$, hence it is coprime with $10$. Maybe you meant the same. I certainly agree that the proof is written too densely. | |
| Apr 20, 2012 at 8:58 | vote | accept | Alexey Ustinov | ||
| Apr 20, 2012 at 8:58 | comment | added | Alexey Ustinov | Very nice arguments. But there is a little gap in the proof. You assume that $b$ is not divisible by $10$, but you use stronger assumption: $(b,10)=1$. I think that we should add some more arguments in this proof. 1) We can start with $a_0$ such that $(a_0,10)=1$ (otherwise we can multiply $a$ by $2^k$ or $5^l$ removing $0$'s). 2) In this case $k$ ($a=11...1=ka_0$) will be coprime with $10$ as well. It means that $b=kb_0$ has no $0$'s at the end. | |
| Apr 20, 2012 at 6:31 | vote | accept | Alexey Ustinov | ||
| Apr 20, 2012 at 6:51 | |||||
| Apr 19, 2012 at 20:42 | comment | added | GH from MO | 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. | |
| Apr 19, 2012 at 20:41 | comment | added | Barry Cipra | Oh sorry! My stupid mistake! | |
| Apr 19, 2012 at 20:40 | comment | added | GH from MO | 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. | |
| Apr 19, 2012 at 20:39 | comment | added | Fedor Petrov | we must have $s(9b)=9s(b)$, for $b=2$ it is not so. | |
| Apr 19, 2012 at 20:34 | comment | added | Barry Cipra | I don't quite follow the "hence $b$ also has only 1's and 0's" conclusion. What about $a=11$ and $b=2$? | |
| Apr 19, 2012 at 20:20 | history | answered | Fedor Petrov | CC BY-SA 3.0 |