whoops sorry - whenever I did it they all came out at divisible only
by 2 or 3. better move on to a new subject, like if you have a prime
of specified value, and create a set s whose members exclusively
comprise it and all primes of lower value, what factor does this prime
have to be multiplied by to always yield a set of different primes
whose cardinality is equal to that of s?
--- In firstname.lastname@example.org, Peter Kosinar <goober@k...> wrote:
> > Sorry - for primes>2, please read primes>3.
> 23 - 13 = 10
> 10 / 2 = 5