Why is $a\equiv b \pmod n$   equivalent to the congruences $a\equiv  b,b+n,b+2n,\dots,b+(c-1)n\pmod {cn}$? I learned the following proposition (in which there is no proof) in a GRE math preparation book. I don't understand what it means and I am not able to find any theorem about this statement in Hardy's An Introduction to the Theory of Numbers.

For any positive integer $c$, the
   statement $a\equiv b \pmod n$ is
   equivalent to the congruences $a\equiv
 b,b+n,b+2n,\dots,b+(c-1)n\pmod {cn}.$

I cannot even apply this proposition to an example such as $7\equiv 1\pmod 6$. If the above is true, then 
$$7\equiv 1,7,13,19\pmod{24}$$
which is obvious not true. 
Is there any typo here? Or how should I understand this "proposition"?
Edit: This question may be related to the question here. 
Added: 

How should I prove this proposition?

 A: Just in case you are not familiar with the equivalence to congruence I am about to use:
Lemma. Let $a$, $b$, and $n$ be integers. Then $a\equiv b\pmod{n}$ if and only if there exists an integer $k$ such that $a=b+kn$.
Proof. $a\equiv b\pmod{n}$ if and only if $n|a-b$, if and only if there exists an integer $k$ such that $nk=a-b$, if and only if there exists an integer $k$ such that $b+nk = a$. QED
To prove the proposition, first assume that $a\equiv b\pmod{n}$. That means that $a=b+kn$ for some integer $k$. Therefore,
$$a\equiv b+kn \pmod{nc}$$
holds. This looks almost like the answer we want. So the question is: what are the possible values for $kn$ modulo $nc$?
To find that out, divide $k$ by $c$ with remainder; that is, write $k=qc+r$, with $0\leq r\lt c$ (division algorithm). Then 
$$b+kn = b+(qc+r)n = b+q(cn) + rn \equiv b+rn\pmod{cn}.$$
Therefore, 
$$a\equiv b+rn\pmod{nc},$$
and $r$ is either $0$, $1$, $2,\ldots,c-1$, because it is the remainder of dividing $k$ by $c$.
Conversely, suppose that 
$$a\equiv b+rn\mod{cn}$$
for some $r$, $r=0$, $1$, $2,\ldots,c-1$. That means that $a=b+rn+k(cn)$ for some integer $k$. Then 
$$a = b+rn+kcn = b+(r+kc)n,$$
so 
$$a =b+(r+kc)n \equiv b\pmod{n}.$$
Thus, $a\equiv b\pmod{n}$ if and only if $a$ is congruent to one of $b$, $b+n$, $b+2n,\ldots,b+(c-1)n$ modulo $cn$.
