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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?

I found the factorization $\,2019 = 3\cdot 673,\,$ but how can that help for a divisibility test?

Bill Dubuque
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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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By CRT = Chinese Remainder Theorem, to compute a remainder $\!\bmod 2019 = 3\cdot 673\,$ it suffices to compute the remainder mod $3\,$ & $\,673,\,$ then apply the CRT formula.

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

$\!\!\bmod 673\!:\ \color{#0af}{10^{\large 14}\equiv 8}\,$ so we can use that to work in chunks of $\,14\,$ decimal digits, e.g.

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

$ $ 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 divisibility test, which is essentially a modular form of the long division algorithm that ignores the quotients.

Alternatively it is the special case $\,\color{#c00}{t = 201}\,$ of a general divisibilit test, viz.

$$\begin{align} n=10\:\!\color{#c00}t\!+\!9&\mid 10b\!+\!a\iff n\mid b\!+\!(1\!+\!\color{#c00}t)a\\[.3em] \underset{\color{#c00}{\!t\,=\,201}}\Longrightarrow n=2019&\mid10b\!+\!a\iff n\mid b\ +\ 202\,a \end{align}\qquad\qquad$$

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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