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Is there exist any known method to find divisibility rule of each and every rational number in any numeral system by analysing its reciprocal. And additionally it will give the remainder on division if not divisible.

What I mean to say is I find it out and want to know that its already known or not?

Like if we want to calculate remainder if $2,183,732,179$ is divided by $37$. First we need to find out the reciprocal of $37$ which is $0.027027027.......$. Now count the number of digits in repeating part which is $3$ digits $027$. Now break the digits of dividend in group of three digits each and add all of them.

$179+732+183+002= 1096$

do it again

$096+001= 97$

Now divide divide $97$ by $37$ we will get same remainder which is $23$.

This is just a simple example, Method is not limited to integers or decimal system.

What I need to know is that method already known or not. If not I need the suggestion which maths/computaional journal may publish the theorem.

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  • $\begingroup$ It is not clear what you mean. Please give a concrete example. $\endgroup$ – Bill Dubuque May 31 '14 at 14:46
  • $\begingroup$ Thanks for suggestion....Suitable example is given $\endgroup$ – Mahendra Goyal Jun 6 '14 at 14:55
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It is an interesting observation, and it is known, though whether it is published in the form you suggest, I don't know. The one-digit version is the old trick of "casting out 9s", and you have generalised this to casting out 99s and 999s. Sometimes this is computationally useful. (note $999=27\times 37$)

Some things people actually use (related to the trick for divisibility by $11$) are connected with $1001$ - a number also beloved of conjurers and tricksters, because $1001=7\times 11\times 13$ deals with the three primes after $2,3,5$ for which there are obvious divisibility tests.

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