If a Carmichael number n has a prime divisor p = 6k+1,

then, since p-1 | n-1 (Korselt's criterion, cf. Wikipedia)

we have that 6k | n-1, such that n = 1 (mod 3).

This means that a Carmichael number cannot be divisible by 3 if it has a prime factor p = 1 (mod 6).

Therefore, if 3 | n, then it cannot have a prime divisor = 1 mod 6.

Maximilian

--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

>

>

>

> --- In primenumbers@yahoogroups.com,

> "mikeoakes2" <mikeoakes2@> wrote:

>

> > Conjecture: If n is a Carmichael number divisible by 3,

> > then all the other prime factors of n are congruent to 5 mod 6.

>

> There are 1967 Carmichael numbers less than 10^18 that

> are divisible by 3 and your conjecture holds for all of them.

>

> David

>