I'm currently working on this problem:
Use the definition of congruence and Fermat's Little Theorem to show that if gcd$(b,561) = 1$, then $$b^{560} \equiv \begin{equation} \begin{cases} 1 &\text{ mod } 3,\\ 1 &\text{ mod } 11,\\ 1 &\text{ mod } 17.\\ \end{cases} \end{equation}$$
Conclude that, if gcd$(b,561)=1$, then $b^{560}\equiv1\text{ mod }561$ and so $561$ is a Carmichael number. Hint: What does it mean that $b^{561}\equiv 1\text{ mod }561?$
So I'm guessing that it's significant that 3,11, and 17 all divide 561. Fermat's little theorem doesn't, by itself, provide for something like $b^{p-1} \equiv 1\text{ mod }q$ where q and p are different, so I'm guessing this is where I need to use congruence to prove that 561 is congruent to 3, 11, and 17?