Are there any primes that are never a factor of a Carmichael number? Is there a prime number $p$ that $p > 2$, and in which $p$ is a never a factor of any Carmichael number $C_n$:
(p ∤ $C_n$)
After a quick glance at some Carmichael number factors, $p$ must be greater or equal to $53$.
 A: The answer to your question should be no. Every odd prime $p$ is a factor of infinitely many Carmichael numbers. To see this let $N$ be a Carmichael number, and $p$ $|$ $N$ with $p$ prime. $p-1$ $|$ $N-1$ is true, then all other prime factors $p_2$, $p_3$.... $p_n$, ($p_2$ * $p_3$*....$p_n$) $=$ $1$ $\pmod {p-1}$. http://www.numericana.com/data/crump.htm says that $\phi(n)$ must be relatively prime to $n$ to be a divisor of a Carmichael number, for when $n$ is prime $\phi(n)$ is $n-1$ and two consecutive integers are coprime. However, some numbers such as $21$, $39$, $55$, and $57$ cannot be divisors of Carmichael numbers. Korselt's criterion defines a Carmichael number $N$ a square free integer such that all primes $p$ $|$ $n$, $p-1$ $|$ $n-1$. Here are simply why these numbers are never divisors of Carmichael Numbers:
Suppose $N$ is a Carmichael Number:
$N$ $|$ $21$ $=$ $3*7$, $N-1$ $|$ $2$, $N-1$ $|$ $2*3$. $N$ $|$ $3$ on the other hand, we also have $N-1$ $|$ $3$, a contradiction since no two consecuative integers are both divisible by a number $>$ $1$.
as for these cases too: 
$N$ $|$ $39$ $=$ $3*13$
$N$ $|$ $55$ $=$ $5*11$
$N$ $|$ $57$ $=$ $3*19$
More simply for that two primes $p$ and $pk+1$, a Carmichael number is never a multiple of $p$($pk+1$).
I hope this helps with some of the understanding of divisors of Carmichael Numbers.
