2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ define law of divisibility $\endgroup$ – clark Apr 12 '13 at 18:39
  • $\begingroup$ @clark for example law of divisibility on 2 is the last digits of number must be even. $\endgroup$ – agustin Apr 12 '13 at 18:41
  • $\begingroup$ so it should be an easier or faster process than doing the actual division ? $\endgroup$ – clark Apr 12 '13 at 18:44
  • $\begingroup$ I guess you mean divisibility criterion, cut-the-knot.org/blue/divisibility.shtml $\endgroup$ – leonbloy Apr 12 '13 at 18:45
  • $\begingroup$ Hint: $\rm 37\mid 10^3 - 1,\ $ or note $\rm\: mod\ 37\!:\ 100\equiv -11,\:$ and $\rm\:{-11}^{-1}\equiv 10.$ $\endgroup$ – Math Gems Apr 12 '13 at 18:46
7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $1000$.

$\endgroup$
3
$\begingroup$

$\rm mod\ 37\!:\ 100^{-1}\!\equiv 10\:$ 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 $

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

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

$\endgroup$
1
$\begingroup$

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 $37$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.