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How can I prove this divisibility rules? $b \in \mathbb{Z}^+$

$2\mid b \Longleftrightarrow 2\mid r_0$

$3\mid b \Longleftrightarrow 3\mid(r_0+r_1+\cdots+r_n)$

$5\mid b \Longleftrightarrow 5\mid r_0$

$9\mid b \Longleftrightarrow 9\mid(r_0+r_1+\cdots+r_n)$

$11\mid b \Longleftrightarrow 11\mid(r_0-r_1+r_2-\cdots+(-1)^n r_n)$

Michael Hardy
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Raphael Tomé Santana
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2 Answers2

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Let

$$b=r_0+(r_1\times 10)+(r_2\times 10^2)+\cdots +(r_n\times 10^n)$$ so since

$$10^k\equiv 0\mod2,\quad\forall k\ge1$$ then $$b\equiv r_0\mod2$$ which prove the first rule and since

$$10^k\equiv 1\mod 3,\quad\forall k\ge0$$ then $$b\equiv r_0+\cdots+r_n\mod3$$ and this proves the second rule.

Can you take it from here to prove the other rules?

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This is not an answer. But I can add some thing in to your list.
Divisibility rules of $7$ and $13.$

1) If $n$ is an integer of the form $ab=10a+b$ where $b$ is an integer between $1$ and $9,$ and $a$ would be any positive integer, then $n$ is divisible by $7$ if and only if $a-2b$ divisible by $7.$
Proof
Let $m=2a-b,$ consider $2n+m=2(10a+b)+(2a-b)=21a.$
Therefore $n$ is divisible by $7$ iff $m$ is divisible by $7.$

2) If $n$ is of the above form $n$ is divisible by $13$ iff $a+4b$ is divisible by $13.$
Proof
Let $m=a+4b,$ consider $4n-m=4(10a+b)-(a+4b)=39a.$
Therefore $n$ is divisible by $13$ iff $m$ is divisible by $13.$

Bumblebee
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