<s>[INST] Solve the following congruence: $x(x+1)(x+2) \equiv 0 \pmod{221}$ 
Find the first five solutions for,
  $$x(x+1)(x+2) \equiv 0 \pmod{221}$$

I am very confused. By CRT,
$x(x+1)(x+2) \equiv 0 \pmod{13}$ and $x(x+1)(x+2) \equiv 0 \pmod{17}$ 
But these two congruences are also reckless.
 [/INST] $$x(x+1)(x+2)\equiv 0\pmod{221}\iff \begin{cases}x(x+1)(x+2)\equiv 0\pmod{13}\\x(x+1)(x+2)\equiv 0\pmod{17}\end{cases}$$
By Euclid's Lemma:
$$\iff \begin{cases}x\equiv \{0,-1,-2\}\pmod{13}\\x\equiv \{0,-1,-2\}\pmod{17}\end{cases}$$
By the Chinese Remainder Theorem (CRT) there are $9$ solutions mod $221$.
E.g., if $x\equiv 0\pmod{13}$ and $x\equiv -1\pmod{17}$, then $x\equiv 169 \pmod{221}$.
You should already know how to use CRT. You'll get:
$$x\equiv \{-2,-1,0,50,51,102,117,168,169\}\pmod{221}$$
</s>