Take any $n \in \mathbb{Z}.$
Then for each such $n$ there is an integer $k$ such that one of the following equations are satisfied. $$n=6k + 0$$ $$n=6k+1$$ $$n=6k+2$$ $$n=6k+3$$ $$n=6k+4$$
$$n=6k+5$$
Can any integer satisfy more than one of the above equalities?
Can any integer NOT satisfy any of of the above equalities?
Then we can describe the equivalence classes of the equivalence relation $\equiv_{6}$ (congruence, mod $6$) as they relate to the division algorithm:
Given any $d \in \mathbb{Z}, d>0$, we know that $\forall n \in \mathbb{Z}$, there exist unique integers $q, r$ such that $n = dq + r$ where $0\leq r< d$.
In this problem, we have the divisor $d = 6$ of a given $n$:
- $k = q\in \mathbb{Z}$ is the unique corresponding quotient which results when dividing $n$ by $d = 6$, and
- $r$ is the unique corresponding remainder, $0\le r < d = 6$, left after dividing that $n$ by $6$.
Then the corresponding equivalence classes can be defined in terms of the the remainder $r$:
$$[r]_6 \in \{[0]_6,[1]_6, [2]_6, [3]_6, [4]_6, [5]_6\} \;\text{ where}$$
$$ [0]_6 = \{...,-12,-6,0,6,12,...\}$$
$$[1]_6 = \{...-11,-5,1,7,13,...\}$$
$$[2]_6 = \{...-10,-4,2, 8, 14...\}$$
$$ \vdots$$
$$[5]_6 = \{...-7,-1,5, 11, 17,...\}$$
For reflextive : $a+b=b+a$ and hence $(a,b)\sim (a,b)$.
symmetric : Let $(a,b)\sim (c,d)\Rightarrow\ a+d=b+c\Rightarrow\ d+a=c+b\Rightarrow (c,d)\sim (a,b)$
transitive : let $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)$, so $a+d=b+c,\ c+f=d+e\Rightarrow\ a+(c+f-e)=b+c\Rightarrow\ a+f=b+e$ and hence $(a,b)\sim (e,f)$. Since $\sim$ is reflexive, symmetric and transitive hence it's an equivalence relation.
Equivalence class of any element $(a,b)=\{(c,d)|a+d=b+c\}$, so can you find three such $(a,b)'s$ such that $12+b=7+a$.
Best Answer
Yes, that's what "congruence mod 5" means; thus $\rm\,[n]_5 = n + 5\,\Bbb Z,\:$ i.e. all $\rm\:k\:$ such that $\rm\:5\:|\:k\!-\!n.$