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What is the divisibility rule or way to find that a number is divisible by 2019 and we can't divide the number by 2019 to test it? and how do we prove that the rule works for 2019?

Help is appreciated!

cookiemonster
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  • $2019 = 3\cdot 673$. – rogerl Aug 08 '19 at 01:09
  • I know but how do we prove that it is the rule for 2019 – cookiemonster Aug 08 '19 at 01:10
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    Well it simplifies the question to showing divisibility by $3$ and $673$ separately at least! – Cheerful Parsnip Aug 08 '19 at 01:12
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    What is "the divisibility rule" from the question? (A rule that we have to search and find first?!) Note that "simple rules" (as for $2,3,4,5,8,9,11$ in basis $10$, obtained by grouping / inspecting the digits) are only possible if the number divides some power of $10$ (as it is the case for $2,4,8,5,25$), or if $10$ to a small power is one modulo the number, as it is the case for $3,9,11$. – dan_fulea Aug 08 '19 at 01:15
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    In our case, the multiplicative order of $10$ in the field $\Bbb F_p$, $p=673$, is $224$, so we could formulate a complicated rule by grouping digits of the given number to be tested for divisibility with $p$ in groups of $224$ digits. You want / need / ask for such a rule?! – dan_fulea Aug 08 '19 at 01:17
  • There does not appear to be any good shortcut for divisibility by $673$. One way to formulate such rules is to take the residue class of $10^n$ modulo $673$ until you find a repeating pattern. However, this will involve a complicated pattern with hundreds of different coefficients! – Cheerful Parsnip Aug 08 '19 at 01:20
  • Thank you for your help – cookiemonster Aug 08 '19 at 01:41

2 Answers2

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$2019 = 3(673)\,$ so it suffices by CRT to compute the remainders mod $3\,$ & $\,673$.

$\!\!\bmod 3\!:\,$ the remainder is congruent to the digit sum (as in casting out nines).

$\!\!\bmod 673\!:\ 10^{\large 14}\equiv 8\,$ so we can use that to work in chunks of $\,14\,$ decimal digits, e.g.

$\ n = 8100000000000025= 81(10^{\large 14})+25 \equiv 81(8)+25\equiv 673\equiv \color{#0a0}0$

$ $ By $ $ Easy $ $ CRT: $\,\ \ \ \begin{align} &n\equiv \color{#0a0}a\pmod{\!673}\\ &n\equiv\color{#c00} b\pmod{\!3}\end{align}\iff\, n\equiv \color{#0a0}a + 673(\color{#c00}b\!-\!\color{#0a0}a)\,\pmod{\!2019}$

e.g. above $\,n\equiv 8\!+\!1+\!2\!+\!5\equiv\color{#c00} 1\pmod{\!3}\,$ so $\ n\equiv \underbrace{\color{#0a0}0+673(\color{#c00}1\!-\!\color{#0a0}0)}_{\large 673}\,\pmod{\!2019}$

For some numbers this may be faster than the universal divsiibility test, which is essentially a modular form of the long division algorithm that ignores the quotients.

Bill Dubuque
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In the spirit of the usual rule for $7$, which checks whether a number is divisible but does not give the remainder if not, given a number $n$ to check, you can delete the last digit of $n$ and add $202$ times that digit to the result. You can keep going until you get down to four digits and there are not many multiples of $2019$ with four digits.

It works because if the last digit is $d$ we are saying that $n$ is divisible by $2019$ if and only if $n+2109d$ is, but $n+2019d$ ends in $10$ so we can divide by $10$.

As an example, if $n=954987$ we next get $$95498+202\cdot 7=96912\\ 9691+202\cdot 2=10095\\ 1009+5\cdot 202=2019$$ so $954987$ is divisible by $2019.$ In fact it is $2019 \cdot 473$

Ross Millikan
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