I'm not sure where you showed the 44 different 5 integer factorizations of 358358?
Thanks, -- Raul On Fri, Nov 3, 2017 at 5:27 AM, Erling Hellenäs <[email protected]> wrote: > Hi all ! > > */q:358358 > 2 7 11 13 179 > */q:358358 > 358358 > q: 40 > 2 2 2 5 > > Which 32 integer factors? > > It seems to be equivalent to the solution I showed? > > We don't have to write factorization programs since it is built in? > > So, it's solved? > > Cheers, > > Erling Hellenäs > > Den 2017-11-03 kl. 08:14, skrev Raul Miller: >> >> Hmm... actually, thinking about it, the par approach here is not >> efficient enough for this example. 5 parRuskeyE 32 is too big of a >> result, I think. (358358 has 5 distinct prime factors and, thus, 32 >> integer factors.) >> >> So, ok, we can do another pass at this problem. >> >> (But, also, this should be a caution about trying to generalize - an >> approach which solves a problem and which solves a generalization of >> that problem will quite often be useless for an instance of a >> different generalization of that problem.) >> >> Thanks, >> > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
