"Had to re-read this twice. ..."


by Alexei Andreev Jun 10 2016

However, it becomes more worth it the more 10\-digits we're splitting\. You might expect that, when splitting three digit wheels, we'd each get one digit wheel plus a 3\-message \(on the third wheel\)\. However, we can actually do better than that\! If you pick a number $~$a$~$ between 0 and 30 \(for a total of 31 different possibilities\), and I pick a number $~$b$~$ between 0 and 30, then the number $~$31a + b$~$ can always be stored on 3 digit wheels \(because it's always less than $~$31\\cdot 30 + 30 \= 960$~$\), so if we're splitting three digit wheels, we can actually eek out one "pseudo 31\-digit" each\. This still isn't maximally efficient \(the values from 961 to 999 are wasted\), but it's a little better\. And if we split five digit wheels, we each get to use one 316\-digit \(as you can verify\)\.

Had to re-read this twice. Not sure if I fully got it.