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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?

Bunny
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Mark
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    Consider the number $5n$ where $n$ is an integer. when $n$ is even, $5n$ is a multiple of $10$ so it ends with digit $0$. Otherwise $5n=10k+5$ for some $k$, so it ends with digit $5$. – user160738 Dec 30 '15 at 01:37
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    @user160738 Is it obvious all multiples of 10 end with digit 0? ;) – Corellian Dec 30 '15 at 02:05

2 Answers2

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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 $5$.

Brian M. Scott
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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) $.

marty cohen
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