Prove every integer is of the form $5k+r$ with $0\le r<5$ I have came across this question from my text book:
Prove or disprove: any integer $n$ is of the form: $5k$, $5k + 1$, $5k + 2$, $5k + 3$ or $5k + 4$ for some integer $k$.
I'm not sure what would be the appropriate method of proof here. Can someone point me in the right direction?
 A: Use the Euclidean Algorithm ( I mean the Division Algorithm ): Given $n$-an interger. Divide $n$ by $5$: $n = 5q+r$, then $r$ as the remainder, it must be non-negative, and less than $5$. Thus: $r = 0,1,2,3,4$. So $n = 5q, 5q+1,5q+2,5q+3$ or $5q+4$, proving the statement.
A: We can prove this for $n\ge 0$ by induction on $n$.  We first check that $n=0=5k$, for $k=0$, as our base case.  Suppose now, our inductive hypothesis, that $n-1=5k+j$, for some $0\le j\le 4$.  If $0\le j\le 3$, then $n=5k+(j+1)$, is of the desired form.  If instead $j=4$, then $n=5(k+1)+0=(5k+4)+1=(n-1)+1$.
I leave those $n<0$ for you to do; they can be done again by induction on $|n|$.  
A: It is a direct consequence of Euclidean division theorem (I've changed it a bit):

Given integers $a,b$ with $b> 0$, there exist integers $q,r$ such that $a=bq+r$ and $0\le r<b$.

Proof: $\lfloor x\rfloor$ is the floor function, i.e. largest integer smaller than or equal to $x$:
$$\lfloor\frac{a}{b}\rfloor\le \frac{a}{b}< \lfloor\frac{a}{b}\rfloor+1$$
$$\iff 0\le \frac{a}{b}-\lfloor\frac{a}{b}\rfloor< 1$$
$$\iff 0\le a-b\lfloor\frac{a}{b}\rfloor< b$$
$$\iff a=b\lfloor\frac{a}{b}\rfloor+r$$
for some $0\le r< b$. QED. So for any $a\in\Bbb Z$ we can find $q,r\in\Bbb Z$ with $0\le r<5$ such that $a=5q+r$.
A: Use long division: any integer $n$ has a quotient $q$ and a remainder $r$, $0\le r<5$, upon division by $5$, and these are unique. 
A: Write the number in decimal and consider the last digit: $n=10m+d$, where $d\in \{0,1,2,\dots,9\}$.
If $d < 5$, then $n=5k+r$, for $k=2m$ and $r=d$.
If $d \ge 5$, then $n=5k+r$, for $k=2m+1$ and $r=d-5$.
