law of divisibility on $37$

Question

how to find and prove law of divisibility on $37$?

Thanks in advance.

Added:---- how to prove for$37$ that: Split off the last digit, multiply by 11, and subtract the product from the number that is left. if the result is divisible by 37 then so is the original number.divisibility criteria for 37

@clark for example law of divisibility on 2 is the last digits of number must be even. — agustin, Apr 12 '13 at 18:41
so it should be an easier or faster process than doing the actual division ? — clark, Apr 12 '13 at 18:44
I guess you mean divisibility criterion, http://www.cut-the-knot.org/blue/divisibility.shtml — leonbloy, Apr 12 '13 at 18:45
Hint: $\rm 37\mid 10^3 - 1,\ $ or note $\rm: mod\ 37!:\ 100\equiv -11,:$ and $\rm:{-11}^{-1}\equiv 10.$ — Math Gems, Apr 12 '13 at 18:46
http://arxiv.org/ftp/math/papers/0001/0001012.pdf might be relevant if you are curious about the first few divisibility rules. — Fixed Point, Apr 12 '13 at 20:20

score 7 · Accepted Answer · edited Apr 13 '17 at 12:20

As $$999=37\cdot27$$

So, $10^3\equiv1\pmod{37}\implies 10^{3k}\equiv1\pmod{37}$

So, $\sum_{0\le r\le n}a_i10^i$

$$=(a_0+10a_1+100a_2)+10^3(a_3+10a_4+100a_5)+10^6(a_6+10a_7+100a_8)\cdots$$

$$\equiv (a_0+10a_1+100a_2)+(a_3+10a_4+100a_5)+(a_6+10a_7+100a_8)\cdots\pmod {37}$$

i.e., we can group by $3$ digits and add to test the divisibility by $37$

EDIT: the explanation of the link in the Question

As $11\cdot(10a+b)-1\cdot(11b-a)=111\cdot a$

$(10a+b)$ will be divisible by $111\iff (11b-a)$ is divisible by $111$

A more general idea can be found here.

Try to find how $11,1$ are identified as multiplier from the link.

score 6 · Answer 2 · answered Apr 12 '13 at 18:45

6

HINT: $37$ divides $999=37\cdot27$, so $37\equiv-1\pmod{1000}$. Now think about the classic divisibility test for $9\equiv-1\pmod{10}$. Imagine writing the number to be tested in base $1000$.

answered Apr 12 '13 at 18:45

Brian M. Scott

616,228

Math Gems · Answer 3 · 2013-04-12T20:13:09.120

$\rm mod\ 37\!:\ 100^{-1}\!\equiv 10\:$ so $ $ e.g. $\rm\:37\mid n = 100^2 c + 100 b + a \iff 37\mid 10^2 n \equiv c + 10 b + 10^2 a$

$37\mid \color{#0a0}{54}\color{blue}{39}\!:\quad \color{#0a0}{54}\!+\!10(\color{blue}{39}) \equiv \color{0a0}{17}\!+\!10(\color{blue}2)\equiv 0 $

$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{44}\color{blue}{47}\color{#c00}{40}\!: &&\!\!\color{#0a0}{44}\!+\!10(\color{blue}{47}\!+\!10(\color{#c00}{40}))\\ \equiv && \color{#0a0}7\!+\!10(\color{blue}{10}\!+\!10(\color{#c00}3))\\ \equiv && \color{#0a0}7\!+\!10(3)\equiv 0\end{eqnarray}$

$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{54}\color{blue}{66}\color{#c00}{65}38\!: &&\color{#0a0}{54}\!+\!10(\color{blue}{66}\!+\!10(\color{#c00}{65}\!+\!10(38))) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10(\color{#c00}{28}\!+\!10)) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10)\\ \equiv &&\color{#0a0}{17}\!+\!10(2)\equiv 0 \end{eqnarray}$

Lord Soth · Answer 4 · 2013-04-12T18:54:44.410

1

Given a number in base 10, say $d_nd_{n-1}\cdots d_0$, calculate $d_0 + 10d_1+26d_2+d_3+10d4+26d_5+\cdots$ (alternating $1$, $10$, and $26$). The resulting number is divisible by 37 if and only if $d_nd_{n-1}\cdots d_0$ is divisible by $37$.

edited Apr 12 '13 at 18:54

answered Apr 12 '13 at 18:48

Lord Soth

7,750
20
37

law of divisibility on $37$

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