ff=: 3 : '(0=+/"(1)7 8 9 0 e.~10 10 10 10 #: y)#y' $ff 1000+i.9000
1296 $ff 1000+i.5667 1296 On Tue, Oct 3, 2017 at 4:54 PM, Roger Hui <[email protected]> wrote: > Why 1000+i.8999? 1000+i.6000 or 1000+i.9000 would make more sense. > > > On Tue, Oct 3, 2017 at 1:48 PM, Raul Miller <[email protected]> wrote: > > > Something like this? > > > > (#~ 1- +./@e.&'7890'@":"0) 1000+i.8999 > > > > Thanks, > > > > -- > > Raul > > > > > > On Tue, Oct 3, 2017 at 4:16 PM, Skip Cave <[email protected]> > wrote: > > > Another interesting Quora problem: > > > > > > Given the integers from 1000 to 9999, remove all integers that contain > a > > 7, > > > 8, 9, or zero, and list the remaining integers. > > > > > > Skip > > > ---------------------------------------------------------------------- > > > For information about J forums see http://www.jsoftware.com/forums.htm > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
