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Given a number presented in decimal form $n=\sum_{k=0} d_k10^k$ there are some nice thumb rules to decide whether it is divisible with a prime number $p\in\{7,11,13,17,19,23,29\}$.

For each of this numbers there is a number $a_p$ such that $p|n\iff p|(a_p\cdot d_0+\sum_{k=0} d_{k+1}10^k)$.

$a_7=-2,\;a_{11}=-1,\;a_{13}=4,\;a_{17}=-5,\;a_{19}=2,\;a_{23}=7,\;a_{29}=3$

E.g. $\;19\cdot 17=323$ and $32+2\cdot 3=38=2\cdot 19$ and $32-5\cdot 3=17$.

Is there a number $a_{31}$ that works the same way? Or $a_p,\;p\ge37$?

Bill Dubuque
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Lehs
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    By this general divisibility test

    $$ \bbox[6px,border:1px solid #c00]{p\mid 10,b+a \iff p\mid b+c,a}\ \ {\rm for}\ \ c\equiv {10}^{-1}!!!!\pmod{!p},\ \ 2,5\nmid p\qquad $$

    e.g. for $,p=29,$ we have $\bmod 29!:\ c\equiv \dfrac{1}{10}\equiv \dfrac{3}{30}\equiv \dfrac{3}1,$ by here. $\ \ $

     – Bill Dubuque Nov 27 '23 at 19:27

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Yes, there is. For all primes $p$ other than $2$ and $5$ the rule $a_p=1/10$ works. The arithmetic is to be done in $\Bbb{Z}_p$ (or, modulo $p$):

  • With $p=7$ we have $10=3$, so $a_7=1/10=1/3=-2$.
  • With $p=11$ we have $10=-1$, so $a_{11}=1/(-1)=-1$.
  • With $p=31$ we have $10\cdot(-3)=-30=1$, so $a_{31}=1/10=-3$.

The reason is that when $10a_p\equiv1\pmod p$ we have $$ n=\sum_{k\ge0}d_k10^k\equiv 10\left(d_0a_p+\sum_{k>0} d_k10^{k-1}\right)\pmod p. $$ Therefore $n$ is divisible by $p$ if and only if $10(d_0a_p+\sum_{k>0}d_k10^{k-1})$ is. Because $\gcd(10,p)=1$ cancelling that factor $10$ won't change divisibility by $p$, and we get that $n$ is divisible by $p$ if and only if the number $$ a_pd_0+\sum_{k>0}d_k10^{k-1}=a_pd_0+\lfloor\frac n{10}\rfloor $$ is divisible by $p$.

Jyrki Lahtonen
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