1

$d\mid10a-1$.
Proof that then $d\mid10q+r$ if and only if $d\mid q+ar$

My attempt was to go once in this direction: $$d\mid10a-1 \wedge d\mid10q+r\implies d\mid q+ar$$ and then with this: $$d\mid10a-1 \wedge d\mid q+ar\implies d\mid10q+r$$ but I only got that: $$d\mid10a-1 \wedge d\mid10q+r$$ then: $$d\mid \left( 10a-1 \right)\left(10q+r\right) + 10q+r$$ $$d\mid 100aq+10ar$$ $$d\mid 100aq+10ar - 9a\left(10q+r\right)$$ $$d\mid 10aq+ar$$ It is somehow "close".

Bill Dubuque
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Piotr Wasilewicz
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    Hint: $q+ar = a (10q+r) - q(10a-1)$ – jjagmath Jul 06 '24 at 12:03
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    Hint: eliminate $10$ from $!\bmod d!:\ \begin{align}10q+r\equiv 0\[.3em] 10a-1\equiv 0\end{align},\ $ i.e. $\ \begin{align} &a,[,10q+r\equiv 0,]\[.2em] -\ &q,[,10a-1\equiv 0,]\[.3em] \hline \Rightarrow &\quad,\ ar,+,q\equiv 0\end{align}.,$ Or, said more arithmetically conceptually: scale $,10q+r\equiv 0,$ by $,10^{-1}\equiv a,,$ cf. linked dupe. $\ \ $ – Bill Dubuque Jul 06 '24 at 18:51
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    $$\begin{align}{\rm Thus}\ \ \ \color{#0a0}{q+ar} = \color{darkorange}a(\color{#0af}{10q+r})-q(\color{#c00}{10a-1})\ \ {\rm so}\ \ d\mid {\rm \color{#0af}{RH\color{#c00}S}}\Rightarrow d\mid{\rm \color{#0a0}{LHS}}\[.3em] {\rm and}\ \ \color{#0af}{10q+r} = \color{darkorange}{10}(\color{#0a0}{q+ar})-r(\color{#c00}{10a-1}) \ \ {\rm so}\ \ d\mid {\rm \color{#0a0}{RH\color{#c00}S}}\Rightarrow d\mid{\rm \color{#0af}{LHS}} \end{align}\qquad$$ where the 2nd equation is derived similarly as the first, by eliminating $,a,$ (vs. $10$), i.e. – Bill Dubuque Jul 06 '24 at 20:23
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    said more arithmetically, by scaling $,\color{#0a0}{q+ar},$ by $10\equiv a^{-1},$ to get $,\color{#0af}{10q+r}.,$ Generally scaling by a unit (invertible), here $,\color{darkorange}{a\equiv 10^{-1}},,$ always yields an equivalent congruence, cf. 2nd dupe. Here it scales between the equivalent congruences $,\bbox[6px,border:1px solid #c00]{\color{#0a0}{q+ar\equiv 0}\overset{\times, \color{darkorange}a}\Leftarrow!!!\underset{\times,\color{darkorange}{10}}\Rightarrow \color{#0af}{10q+r\equiv 0},\pmod{!d}}\ \ $ – Bill Dubuque Jul 06 '24 at 20:24

2 Answers2

1

$d\mid10a-1\;.$
Prove that $\;d\mid10q+r\iff d\mid q+ar\;.$

Proof$\,\implies:$
Since $\;d\mid10a-1\,,\;$ it follows that $\;d\mid r(10a-1)\;.$
Since $\;d\mid r(10a-1)\;$ and $\;d\mid 10q+r\;,\;$ we get that
$d\mid r(10a-1)+10q+r\;$ that is $\;d\mid10(q+ar)\;.$
Since $\;d\mid10a-1\,,\;$ it follows that $\;2\not\mid d\;$ and $\;5\not\mid d\;,\;$ hence
$GCD(d,10)=1\;.$
Since $\;d\mid10(q+ar)\;$ and $\;GCD(d,10)=1\;,\;$ we get that
$d\mid q+ar\;.$

Proof$\,\impliedby:$
Since $\;d\mid10a-1\,,\;$ it follows that $\;d\mid q(10a-1)\;.$
Since $\;d\mid q(10a-1)\;$ and $\;d\mid q+ar\;,\;$ we get that
$d\mid q(10a-1)+q+ar\;$ that is $\;d\mid a(10q+r)\;.$
Since $\;d\mid10a-1\,,\;$ it follows that there not exists any $\,p\,$ prime such that $\;p\mid d\;$ and $\;p\mid a\;,\;$ hence
$GCD(d,a)=1\;.$
Since $\;d\mid a(10q+r)\;$ and $\;GCD(d,a)=1\;,\;$ we get that
$d\mid10q+r\;.$

Angelo
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jul 06 '24 at 17:11
  • See my comments on the question for the arithmetical idea behind such equivalences (unit scaling equivalence). This makes proofs like the above short and trivial, and conceptually much clearer. – Bill Dubuque Jul 06 '24 at 20:32
0

Given that, $d \mid (10a -1)$. Now as you have shown, $d \mid 10aq + ar$ so $d \mid (10aq + ar) - q(10a -1) \implies d \mid q + ar$

If $d \mid (q + ar)$ then $d \mid 10(q+ar) - r(10a -1) \implies d \mid 10q+r$.

Afntu
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jul 06 '24 at 17:11