This should work Skip, although I changed your call to q: to work on scalars and work with a nested list …
q: each 1998 2997 3996 4995 5994 6993 7992 8991 ┌──────────┬──────────┬────────────┬──────────┬────────────┬──────────┬──────────────┬────────────┐ │2 3 3 3 37│3 3 3 3 37│2 2 3 3 3 37│3 3 3 5 37│2 3 3 3 3 37│3 3 3 7 37│2 2 2 3 3 3 37│3 3 3 3 3 37│ └──────────┴──────────┴────────────┴──────────┴────────────┴──────────┴──────────────┴────────────┘ ~. each q: each 1998 2997 3996 4995 5994 6993 7992 8991 ┌──────┬────┬──────┬──────┬──────┬──────┬──────┬────┐ │2 3 37│3 37│2 3 37│3 5 37│2 3 37│3 7 37│2 3 37│3 37│ └──────┴────┴──────┴──────┴──────┴──────┴──────┴────┘ in=:(([ e. ])#[) 2 3 37 in 3 37 3 37 in each/ ~. each q: each 1998 2997 3996 4995 5994 6993 7992 8991 ┌────┐ │3 37│ └────┘ …/Rob > On 23 Oct 2018, at 5:31 pm, Skip Cave <[email protected]> wrote: > > Given the integers: 1998 2997 3996 4995 5994 6993 7992 8991 > Find the common prime factors in all of these integers. > > Obviously we can find the prime factors of each of the integers: > > q: 1998 2997 3996 4995 5994 6993 7992 8991 > > 2 3 3 3 37 0 0 > > 3 3 3 3 37 0 0 > > 2 2 3 3 3 37 0 > > 3 3 3 5 37 0 0 > > 2 3 3 3 3 37 0 > > 3 3 3 7 37 0 0 > > 2 2 2 3 3 3 37 > > 3 3 3 3 3 37 0 > > > What J expression will find the common factors in all 6 of these integers? > > (the result of the expression should be that there are two common factors - > 3 & 37) > > 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
