Timeline for Equal digit sums
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2012 at 8:58 | vote | accept | Alexey Ustinov | ||
| Apr 20, 2012 at 6:31 | vote | accept | Alexey Ustinov | ||
| Apr 20, 2012 at 6:51 | |||||
| Apr 19, 2012 at 20:20 | answer | added | Fedor Petrov | timeline score: 18 | |
| Apr 19, 2012 at 18:58 | comment | added | Kevin Buzzard | 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 $a=b$. | |
| Apr 19, 2012 at 15:48 | comment | added | Noam D. Elkies | [that is, we may assume without loss of generality $a>b$; if $\gcd(a,10)=1$ then there exists $n$ etc.] | |
| Apr 19, 2012 at 14:51 | comment | added | Noam D. Elkies | 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 $5$. | |
| Apr 19, 2012 at 13:40 | comment | added | Noam D. Elkies | 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 $10$. | |
| Apr 19, 2012 at 13:30 | comment | added | Woett | 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 $3$. | |
| Apr 19, 2012 at 13:25 | comment | added | Woett | Maybe the following properties of the function $s$ could be useful; $s(a+b) \le s(a) + s(b)$ and $s(a)s(b) \le s(a)s(b)$. | |
| Apr 19, 2012 at 13:15 | history | asked | Alexey Ustinov | CC BY-SA 3.0 |