Is there a mathematical proof that shows all multiples of $5$ either end with a $0$ or $5$? I know that all multiples of $5$ end up with a $0$ or $5$ as the last digit. But there are an infinite amount of numbers. Is there a way to formally prove that this is true for all numbers using variables?
 A: Let $n$ be a multiple of $5$, say $n=5m$ for some integer $m$. If $m$ is even, there is an integer $k$ such that $m=2k$, and then $n=10k$. If, on the other hand, $m$ is odd, there is an integer $k$ such that $m=2k+1$, and in that case $n=10k+5$. To complete the argument, we need only show that every multiple of $10$ ends in $0$.
Suppose that $n$ is a multiple of $10$, and suppose that when written in ordinary decimal notation, it is $d_rd_{r-1}\ldots d_0$, where the $d_k$ are the digits. Then
$$\begin{align*}
n&=10^rd_r+10^{r-1}d_{r-1}+\ldots+10d_1+d_0\\
&=10\left(10^{r-1}d_r+10^{r-2}d_{r-1}+\ldots+10d_2+d_1\right)+d_0\;,
\end{align*}$$
where the quantity in parentheses is an integer. Thus,
$$d_0=n-10\left(10^{r-1}d_r+10^{r-2}d_{r-1}+\ldots+10d_2+d_1\right)\;,\tag{1}$$
and if $n$ is a multiple of $10$, the righthand side of $(1)$ is a multiple of $10$. Thus, $d_0$ is a multiple of $10$. But $0\le d_0\le 9$, so $d_0=0$. This proves that every multiple of $10$ ends in $0$ and hence that every multiple of $5$ ends in $0$ or $5$.
A: Suppose $n = 10m+k$,
$0 \le k \le 9$,
is not a multiple of $5$.
Since $10m = 2\cdot 5m$
is a multiple of $5$,
$k$ is not a multiple of $5$.
Therefore
$k$ can not be $0$ or $5$,
but can be any other digit.
Therefore the last digit
being $0$ or $5$
is necessary for
$n$ to be a multiple of $5$.
To show that the
last digit being
$0$ or $5$
is sufficient to make $n$
a multiple of $5$,
write
$10m+0 = 2\cdot 5 m
=5(2m)
$
and
$10m+5 = 2\cdot 5 m+5
=5(2m+1)
$.
