Find number of invertible elements in $\mathbb{Z}[i]/(220+55i)\mathbb{Z}[i]$ I was able to find the factorization
$220+55i=11*(2+i)*(2-i)*(4+i)$.
Also know this famous question 
Quotient ring of Gaussian integers
But how to apply it in this case? I'm confused, please help.
Is the answer
$$
(11^2-1)*(5-1)*(5-1)*(17-1)
$$
correct? Why?
 A: As I mentioned in the comments, since the ideals generated by the elements of your factorization are comaximal, then
\begin{align*}
\frac{\mathbb{Z}[i]}{(220+55i)} = \frac{\mathbb{Z}[i]}{(11)(2+i)(2-i)(4+i)} \cong \frac{\mathbb{Z}[i]}{(11)} \times \frac{\mathbb{Z}[i]}{(2+i)} \times \frac{\mathbb{Z}[i]}{(2-i)} \times \frac{\mathbb{Z}[i]}{(4+i)}
\end{align*}
by the Chinese Remainder Theorem. Now
$$
\frac{\mathbb{Z}[i]}{(11)} \cong \frac{\mathbb{Z}[x]/(x^2+1)}{(11, x^2+1)/(x^2+1)} \cong \frac{\mathbb{Z}[x]}{(11, x^2 + 1)} \cong \frac{\mathbb{F}_{11}[x]}{(x^2+1)} \cong \mathbb{F}_{11^2}
$$
by the Third Isomorphism Theorem, and because $x^2+1$ is irreducible over $\mathbb{F}_{11}$. ($\mathbb{F}_q$ denotes the finite field with $q$ elements.) Thus by the result in your linked post we have
$$
\frac{\mathbb{Z}[i]}{(220+55i)} \cong \mathbb{F}_{11^2} \times \mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_{17} \, .
$$
Thus an element of ${\mathbb{Z}[i]}/{(220+55i)}$ is a unit iff its image in each of the factors of $\mathbb{F}_{11^2} \times \mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_{17}$ is a unit. Since every element of $\mathbb{F}_q$ is a unit except for $0$, the answer you've given follows.
