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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?

Stefan4024
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B. David
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    I'd add a step of $-x+30a_0$ so it's clear what you did there. – Dan Uznanski Dec 03 '17 at 14:23
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    @David, See : https://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Dec 04 '17 at 09:19
  • Attention I only notice that 30=2x3x5 IS a primorial. (2,6,30,210,2310,....) Maybe it is not only a coincidence. I don't know. Maybe see :https://math.stackexchange.com/questions/4131235/a-divisibility-rule-for-7-in-primoradic-is-that-already-known – Stéphane Jaouen May 09 '21 at 09:17

1 Answers1

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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 $43$

Stefan4024
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