I think I'd do it like this: #10#.(#~ 36=*/@|:) 1+(6#3)#:i.3^6 90
In other words, generate all 729 possibilities as lists of digits and keep those whose product is 36. That said, technically we don't actually need to combine them into single numbers: #(#~ 36=*/@|:) 1+(6#3)#:i.3^6 90 Thanks, -- Raul On Mon, Apr 9, 2018 at 1:51 AM, Skip Cave <s...@caveconsulting.com> wrote: > Here's a fun math challenge on Quora: > > Find the number of 6-digit numbers made up of the digits 1, 2, and 3 which > have no digit recurring three or more times, consecutively? > > The link to my answer on Quora is: https://goo.gl/BzBDQe > > Skip Cave > Cave Consulting LLC > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
