3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $43$

$\endgroup$
  • $\begingroup$ But what happen with $13a_0$? $\endgroup$ – B. David Dec 3 '17 at 19:14
  • 1
    $\begingroup$ $13a_0 \equiv 13 a_0 - 43a_0 \equiv -30a_0 \pmod{43}$, right? $\endgroup$ – Stefan4024 Dec 3 '17 at 19:19
  • $\begingroup$ Thanks for your help :) $\endgroup$ – B. David Dec 3 '17 at 19:22
  • 1
    $\begingroup$ @David You're welocme! That's why we're here after all $\endgroup$ – Stefan4024 Dec 3 '17 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.