I'm looking for a divisibility test for numbers of the form $10n \pm1$

I need the test to be a summation across digits like the one for $11$, that being, a number $\overline{d_n \ldots d_1 }$ is divisible by 11 if $$ \sum_{i \; even} d_{i+1} - d_i \pmod{11} = 0 $$

I'm after a similarly styled test for all $10n \pm1$. Is there a nice way to generate them?

Thanks in advance for your help!

  • 1
    $\begingroup$ @Peter Neither. I want a rule for 19, 21, 29, 31, 39, 41, ... $\endgroup$ – Ben Crossley Jun 18 '19 at 7:51
  • $\begingroup$ Sorry, I somehow read $10^n$ instead of $10n$ $\endgroup$ – Peter Jun 18 '19 at 7:52
  • $\begingroup$ @Peter So then $10n \equiv -9n \pmod{19}$ would give us sum of even digits minus 9 times the odd digits? Or $100n \equiv 5n \pmod{19}$ So 10 times even + 5 times odd digits. $\endgroup$ – Ben Crossley Jun 18 '19 at 8:04
  • $\begingroup$ The divisibility rule for $11$ only works because the residue of $10^n$ switches between $-1$ and $1$. For $19$, the rule is much more complicated. $\endgroup$ – Peter Jun 18 '19 at 8:09
  • $\begingroup$ I see, is that why the rule for 7 involves 142857? I noticed that 10^6-1 =999999, and 999999/7 = 142857. I'm assuming it's to do with the first 10^n-1 that the divisor goes into. $\endgroup$ – Ben Crossley Jun 18 '19 at 8:12

Example for the number $19$ : The order of $10$ modulo $19$ is $18$, therefore you can divide the given number into $18$-digit-blocks from behind. The first block usually will have less than $18$ digits. Then, you can sum up the residues modulo $19$ of those blocks (considered as a natural number) and the given number is divisible by $19$ if and only if the sum is divisible by $19$.

A bit easier to use is the following rule : Divide the number into $9$-digit blocks from behind, the first block usually will have less than $9$ digits. Then, begin with residue modulo $19$ of the first block, then subtract the residue of the next, then add and so on. The given number is divisible by $19$ , if and only if the final result is divisible by $19$.

Example : $$5645012950185238747288629$$

Dividing gives $$5645012\ 950185238747288629$$

$5645012$ has redisue $17$ and $950185238747288629$ has redisue $2$ , the sum is $19$. Hence , the given number is disible by $19$

$$5645012\ 950185238\ 747288629$$

If we divide in $9$-digit - blocks, we get the residues $17,7,9$ and $17-7+9=19$ which is divisible by $19$, hence the given number is disivible by $19$.

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    $\begingroup$ Thank you. Do you know of a method for 1-digit blocks / if one exists? $\endgroup$ – Ben Crossley Jun 18 '19 at 10:13
  • $\begingroup$ Within the $18$-digit blocks and with the suitable "weights". we could construct a formula using the single digits, but this would even be more messy and moreover, we would not have considered the first block. $\endgroup$ – Peter Jun 18 '19 at 10:15
  • $\begingroup$ An analogue to $11$ does not exist. $\endgroup$ – Peter Jun 18 '19 at 10:17

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