Editing Divisibility rule (section)

==Generalized divisibility rule==

To test for divisibility by ''D'', where ''D'' ends in 1, 3, 7, or 9, the following method can be used.<ref>Dunkels, Andrejs, "Comments on note 82.53—a generalized test for divisibility", ''[[Mathematical Gazette]]'' 84, March 2000, 79–81.</ref> Find any multiple of ''D'' ending in 9. (If ''D'' ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as ''m''. Then a number ''N'' = 10''t'' + ''q'' is divisible by ''D'' if and only if ''mq'' + ''t'' is divisible by ''D''. If the number is too large, you can also break it down into several strings with ''e'' digits each, satisfying either 10<sup>''e''</sup> = 1 or 10<sup>''e''</sup> = −1 (mod ''D''). The sum (or alternating sum) of the numbers have the same divisibility as the original one.

For example, to determine whether 913 = 10 × 91 + 3 is divisible by 11, find that ''m'' = (11 × 9 + 1) ÷ 10 = 10. Then ''mq'' + ''t'' = 10 × 3 + 91 = 121; this is divisible by 11 (with quotient 11), so 913 is also divisible by 11. As another example, to determine whether 689 = 10 × 68 + 9 is divisible by 53, find that ''m'' = (53 × 3 + 1) ÷ 10 = 16. Then ''mq'' + ''t'' = 16 × 9 + 68 = 212, which is divisible by 53 (with quotient 4); so 689 is also divisible by 53.

Alternatively, any number ''Q'' = 10''c'' + ''d'' is divisible by ''n'' = 10''a'' + ''b'', such that gcd(''n'', 2, 5) = 1, if ''c'' + ''D''(''n'')''d'' = ''An'' for some integer ''A'', where
<math display="block">
 D(n) \equiv \begin{cases}
  9a + 1 & \text{if } n = 10a + 1, \\
  3a + 1 & \text{if } n = 10a + 3, \\
  7a + 5 & \text{if } n = 10a + 7, \\
  a + 1  & \text{if } n = 10a + 9.
\end{cases}
</math>

The first few terms of the sequence, generated by ''D''(''n''), are 1, 1, 5, 1, 10, 4, 12, 2,&nbsp;... {{OEIS|A333448}}.

The piece wise form of ''D''(''n'') and the sequence generated by it were first published by Bulgarian mathematician Ivan Stoykov in March 2020.