# Divisibility rule for 43

Let $a_ka_{k-1}\dots a_1a_0$ the decimal expression of number ${n}$. Prove $n$ is divisible by 43 if and only if $a_ka_{k-1}\dots a_1-30a_0$ is divisible by 43.

Proof:

Let $\boldsymbol{x=a_ka_{k-1}\dots a_1}$ and $\boldsymbol{m=x-30a_0}$ then:

\begin{split} 43|n =43 \,|\, 10x+a_0 \Leftrightarrow & 10x&+&a_0 &\equiv 0\ ( \textrm{mod 43)} \\ \Leftrightarrow & 50x&+&5a_0 &\equiv0 \ (\text{mod 43)} \\ \Leftrightarrow & 7x&+&5a_0 &\equiv0 \ (\text{mod 43)} \\ \Leftrightarrow & 42x&+&30a_0 &\equiv0 \ (\text{mod 43)} \\ \Leftrightarrow & x &-& 30a_0& \equiv0 \ (\text{mod 43)} \Leftrightarrow 43 |x-30a_0 \Leftrightarrow 43|m \end{split}

Is correct my proof ? Is there a better proof?

You're proof is perfectly fine. Maybe faster way to prove it to multiply everything by $13$ in the first step. So you have:
$$10x + a_0 \equiv 0 \pmod{43} \iff 130x + 13a_0 \equiv 0 \pmod{43} \iff x - 30a_0 \equiv 0 \pmod{43}$$
If you wonder how we came up with $13$ note that $10 \cdot 13 \equiv 1 \pmod {43}$, so $13$ is the multiplicative inverse of $10$ modulo $43$
• But what happen with $13a_0$? – B. David Dec 3 '17 at 19:14
• $13a_0 \equiv 13 a_0 - 43a_0 \equiv -30a_0 \pmod{43}$, right? – Stefan4024 Dec 3 '17 at 19:19