State the Chinese Remainder Theorem.

A composite number $n$ is defined to be a Carmichael number if $b^{n-1} \equiv 1 \bmod n$ whenever $(b, n)=1$. Show that a composite $n$ is Carmichael if and only if $n$ is square-free and $(p-1)$ divides $(n-1)$ for all prime factors $p$ of $n$. [You may assume that, for $p$ an odd prime and $\alpha \geqslant 1$ an integer, $\left(\mathbb{Z} / p^{\alpha} \mathbb{Z}\right)^{\times}$is a cyclic group.]

Show that if $n=(6 t+1)(12 t+1)(18 t+1)$ with all three factors prime, then $n$ is Carmichael.