Timeline for The digit sum: $s(na)=s(nb)$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2012 at 7:24 | vote | accept | Seva | ||
| Apr 20, 2012 at 21:28 | history | edited | fedja | CC BY-SA 3.0 |
added 1148 characters in body; added 3 characters in body; added 2 characters in body; edited body
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| Apr 20, 2012 at 15:25 | comment | added | Fedor Petrov | @Seva: if $n=10^{M+k}x+y$, where $y$ has less then $k$ digits, and $a$ less then $M$ digits, then $s(na)=s(xa)+s(ya)$. Apply this for all portions of consecutive zeros Fedja is telling about. | |
| Apr 20, 2012 at 14:40 | comment | added | Seva | Sorry - I cannot follow, starting from the sentence "$M$ zeroes will control the transfers ...". If you happen to feel a little more serious about your answer, would you kindly try to add some explanations? | |
| Apr 20, 2012 at 13:26 | comment | added | Emil Jeřábek | Claim 3 was the point of the post Seva referred to. | |
| Apr 20, 2012 at 13:07 | history | answered | fedja | CC BY-SA 3.0 |