Timeline for Decimal digits multiplied by powers of 2: leads to mod 8?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2014 at 12:41 | comment | added | Lev Borisov | I think the same argument will work for any $b_1,b_2$. You will get the same result modulo $b_1-b_2$, but can never get a number less than $b_2$ unless you start with it. | |
| Jul 6, 2014 at 12:34 | vote | accept | Joseph O'Rourke | ||
| Jul 6, 2014 at 12:29 | history | answered | Lev Borisov | CC BY-SA 3.0 |