Verifying Carmichael numbers I'm trying to understand a solution I was given in a tutorial regarding a problem with Carmichael numbers and I was wondering if you guys can help clarify things:

A composite number $m$ is called a Carmichael number if the congruence $a^{m-1} \equiv 1 \pmod{m}$ is true for every number a with $\gcd(a,m) = 1$. 

Verify that $m = 561 = 3 \times 11 \times 17$ is a Carmichael number.

Solution given:

Apply Fermat's Little Theorem to each prime divisor of $m$:
\begin{align*}
a^2 &amp;\equiv 1 \pmod{3}\\
a^{10} &amp;\equiv 1 \pmod{11}\\
a^{16} &amp;\equiv 1 \pmod{17}
\end{align*}
This somehow then implies that $a^{80} \equiv 1 \pmod{561}$ then accordingly $a^{560} \equiv 1 \pmod{561}$.

I am lost as to how the 3 congruences imply $a^{80} \equiv 1 \pmod{561}$ ($80 = \mathrm{LCM}(2,10,16)$).

Can somebody clarify this for me? 

Thanks!
Note that $80 = \mathrm{lcm}(2,10,16)$. So you can write $a^{80} = (a^2)^{40} = (a^{10})^{8} = (a^{16})^5$. So, 
\begin{align*}
a^{80}= (a^2)^{40} &amp;\equiv 1^{40} = 1\pmod{3},\\
a^{80}= (a^{10})^{8} &amp;\equiv 1^8 = 1 \pmod{11},\\
a^{80}= (a^{16})^5 &amp;\equiv 1^5 = 1\pmod{17}.
\end{align*}

By the Chinese Remainder Theorem, the system of congruences 
\begin{align*}
x&amp;\equiv 1\pmod{3}\\
x&amp;\equiv 1\pmod{11}\\
x&amp;\equiv 1\pmod{17}
\end{align*}
has a unique solution modulo $3\times 11\times 17 = 561$. But both $x=1$ and $x=a^{80}$ are solutions. Since the solution is unique modulo $561$, then 
the two solutions we found must be congruent. That is,
$$a^{80}\equiv 1\pmod{561}.$$

(Added. Or, more simply, as Andres points out, since $3$, $11$, and $17$ each divide $a^{80}-1$, and are pairwise relatively prime, then their product divides $a^{80}-1$). 

Once you have that $a^{80}\equiv 1\pmod{561}$, then any power of $a^{80}$ is also congruent to $1$ modulo $561$. In particular,
$$a^{560} = (a^{80})^{7} \equiv 1^7 = 1 \pmod{561}$$
as desired.
You will have to consider two cases though. Case 1 will be that $a$ is not relatively prime to $561$ and case two will be that $5$61 and $a$ are relatively prime.