3 ms·

From: http://quaxio.com/triple/ http://quaxio.com/triple/ ^([0369]|[258][0369][147]|[147]([0369]|[147][0369][258])[258]|[258][0369][258]([0369]|[147][0

by dudus 13y ago

From: http://quaxio.com/triple/ http://quaxio.com/triple/ ^([0369]|[258][0369]*[147]|[147]([0369]|[147][0369]*[258])*[258]|[258][0369]*[258]([0369]|[147][0369]*[258])*[258]|[147]([0369]|[147][0369]*[258])*[147][0369]*[147]|[258][0369]*[258]([0369]|[147][0369]*[258])*[147][0369]*[147])*$

moron4hire 13y ago
569pts: ([^31]0|31|[017]2|[03]03|[^1]4|(900|01|7)5|6|[48]7|[57]8|09)$
- chch 13y ago
  Originally had 568, then saw this and improved. :) 580pts: 00(0$|3|6|9|12|15)|[^0]14|.53|^3[^38]|55|43|23|9.7
  - 3JPLW 13y ago
    Nice. A smidge better at 582: 5[54]|2[437]|00($|[369]|1[25])|^8[17]|^3[29]|9.7
    - jaytaylor 13y ago
      One more smidgen, behold 584: 5[54]|2[437]|00($|[369]|1[25])|^[83][1729]|9.7
      - grobie 13y ago
        586: ^[378][12479]|00($|[369]|1[25])|5[45]|2[347]
        nwellnhof 13y ago
        589: ^[378][12479]|00($|[369]|1[25])|55|2[347]
raverbashing 13y ago
The idea is as thus: Remember division rules? If the sum of digits is divisible by 3, then the number is divisible by 3 So, if you have 0, 3, 6, 9, it's as you can remove them, the value won't change Remaining are the 2, 3, etc digit numbers divisible by 3