q:+./1998 2997 3996 4995 5994 6993 7992 8991
3 3 3 37
Den tirsdag den 23. oktober 2018 08.45.10 CEST skrev 'Rob Hodgkinson' via
Programming <[email protected]>:
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
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