# Modular arithmetic: confusion one the $mod$ operator.

Suppose I have $a\equiv b \text{ (mod$c$)}$ and I just know $c$. I want to know, say $b$. Is this the same as $a$ mod $c$? If so, why?

I think I confuse the congruence with the equality symbol, because only recently I learned that sometimes $mod$ is used as an operator, and not as a mere part of the congruence symbol.

One of the possible values for $b$ would be $a$ mod $c$, but there will be others too, because adding any multiple of $c$ to a solution yields another solution.
For example: $6 \equiv b$ (mod 3). We have (6 mod 3) = 0, and one of the possible values for $b$ is 0, but actually $b$ could be any multiple of 3.
Actually $a\equiv b\text{ ( mod$c$)}$ means $a-b=ck$, where $k\in\mathbb{Z}$. That is when a and b are divided by c in both cases we have same remainder.