Q:

Complete System of Residues modulo $m=m_1m_2...m_r$

Q: Suppose that $m_1,m_2,...,m_r$ are pairwise relatively prime positive integers. For each $j$, let $f(m_j)$ denote a complete system of residues modulo $m_j$.Show the the numbers
$$c_1+c_2m_1+...+c_rm_1m_2...m_{r-1}$$
form a complete system of residues modulo $m=m_1m_2...m_j$, where $c_j \in f(m_j)$.
My Thought:
I nearly have no idea to do this question. Someone show me an answer of this question. He firstly assume
$$c_1+c_2m_1m_2+...+c_rm_1m_2...m_{r-1} \equiv d_1+d_2m_1m_2+...+d_rm_1m_2...m_{r-1} \quad \text{mod } m$$
$$(c_1-d_1)+(c_2-d_2)m_1m_2+...+(c_r-d_r)m_1m_2...m_{r-1} \equiv 0 \quad \text{mod } m$$
Then saying $m_1|c_1-d_1$ so that $c_1=d_1$. Further to continue, divides the congruence equation by $m_1$. (I understand this step as $m$ are relatively prime, dividing $m_1$ is a possible operation). Then continue the above procedures with $c_2$ until $c_r$.
can anyone help me to explain more in detail what it is going on? Or what is the target in the questions. Thank You.

A:

You added an extra $m_2$ at the second term, which should be $c_2m_1$ (note that the $i$th term does not contain $m_i$).
We write $a\equiv b\pmod{n}$ to mean that $a$ and $b$ give the same remainder modulo $n$, i.e. $n\,|\,a-b$.
The proof you wrote then proves that there is no repetition among the values modulo $m$. 
Indeed, if $\ c_1+c_2m_1+c_3m_1m_2+\dots+c_rm_1m_2\dots m_{r-1}\ \equiv\ d_1+d_2m_1+\dots+d_rm_1m_2\dots m_{r-1} \pmod m$, 
then $m\,|\ (c_1-d_1)+\,(\dots)m_1$, hence indeed $m_1|\,c_1-d_1$, i.e. $c_1\equiv d_1\pmod{m_1}$ which now implies $c_1=d_1$ as they from a system of residues.
Going forward, we have $m\,|\ (c_1-d_1)+(c_2-d_2)m_1+\,(...)m_1m_2\ =(c_2-d_2)m_1+\,(...)m_1m_2$, so in particular
$$m_2\,|\ (c_2-d_2)m_1$$
which implies $m_2\,|\ (c_2-d_2)$ because $\gcd(m_1,m_2)=1$.
And, so on. In the end, we will arrive to $c_i=d_i$ for all $i$.
Finally, either one can count the numbers arising with the different $c_i$'s, or we can directly show that every residue modulo $m$ is represented:
Let $a\in\Bbb Z$ be any integer, we can explicitly find the $c_i$ 'coordinates' for it: 
There's a unique $c_1\in f(m_1)$ such that $a\equiv c_1\pmod{m_1}$.
Now, as $m_1$ is coprime to $m_2$, there is a unique $c_2$ such that $c_2m_1\equiv(c_1-a)\pmod{m_2}$.
And so on..

