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I was reading about a not so practical way to determine the divisibility of a number by $7$.
At some point the following number is mentioned: $142857$ (which is the result of $\frac{999999}{7}$) and apparently this number as I have verified if multiplied by $2,3,4,5, 6$ it gives the same digits in different order e.g. $142857 \cdot 3 = 428571$

Why does this number have this property? I see that the numbers $2,3,4,5,6$ are all remainders if we divide $100, 10, 10000, 100000, 1000$ by $7$ respectively but I am not sure if there is any correlation with the property.
I'd like to understand the intuition behind this "trick"

Note: I have found a similar post but I don't see any explanation on this

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  • $\begingroup$ $$\frac{142857}{999999}=\frac17.$$ So $$\frac17=0.\overline{142857}.$$ Then use what you noticed. $\endgroup$ Sep 21 at 21:54
  • $\begingroup$ More generally if $10$ is a generator modulo prime $p,$ you will get something like this. For example, $p=17$ gives the same result for $058235294117647,$ multiplying by $1,\dots,16.$ It is not just a different order, but a rotated order - some number of digits are cut off the end of $142857$ and placed in the front. $\endgroup$ Sep 21 at 22:04
  • $\begingroup$ See en.wikipedia.org/wiki/Cyclic_number $\endgroup$
    – lhf
    Sep 21 at 22:12
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    $\begingroup$ Does this answer your question? Why $142857$ when multiplied by $1$ to $6$ gives the same digits? $\endgroup$
    – Jean Marie
    Sep 21 at 23:57
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    $\begingroup$ @JeanMarie Same question yes , but the answers are so useless. I think we can get a better dupe target. $\endgroup$ Sep 22 at 8:13
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Well, one can note that $142857$ is a bit of a special number in that $\frac{1}{7}=0.\overline{142857}$, which means that, by long division, one obtains $\overline{142857}=142857142857142857...$ ad infinitum via the following algorithm:

Start with $1$.

$1=\color{green}{0}\times7+\color{red}{1}$ so $\frac{1}{7}=0.\text{something}$

To get this something, do:

$\color{red}{1}\times10 = 10 = \color{green}{1}\times7+\color{red}{3}$ (*)

$\color{red}{3}\times10 = 30 = \color{green}{4}\times7+\color{red}{2}$ (**)

$\color{red}{2}\times10 = 20 = \color{green}{2}\times7+\color{red}{6}$

$\color{red}{6}\times10 = 60 = \color{green}{8}\times7+\color{red}{4}$

$\color{red}{4}\times10 = 40 = \color{green}{5}\times7+\color{red}{5}$

$\color{red}{5}\times10 = 50 = \color{green}{7}\times7+\color{red}{1}$

Now, the remainder is $\color{red}{1}$, which is what we started with at equation (*), so we now have a loop which yields $\frac{1}{7}=0.\color{green}{\overline{142857}}$. In group-theoretic parlance, the fact that the loop closes after $6$ iterations means that $10$ has order $6$ in $(\mathbb{Z}/7\mathbb{Z})^*$, i.e. $10^6\equiv1 \pmod{7}$ and $10^k\neq1\pmod7$ for $1 \leq k \leq 5$.

To understand why the digits are shifted when e.g. multiplying by $3$, just note that multiplying $\frac{1}{7}$ by $3$ means considering $\frac{3}{7}$ instead of $\frac{1}{7}$, and so the algorithm would start at equation (**) instead of (*), thus yielding $\color{green}{\overline{428571}}$ instead of $\color{green}{\overline{142857}}$.

So $\frac{3}{7}=\frac{3\times 142857}{999999} = \frac{428571}{999999} \Rightarrow 3\times142857 = 428571$

This explains the shift.

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  • $\begingroup$ I don't follow the logic from the To get this something, do and on. What exactly does the remainder with 10,30,20,60,40,50 show and what is the idea behind doing this to determine the decimal part of the result of $\frac{1}{7}$? $\endgroup$
    – Jim
    Sep 22 at 16:36
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Good question!

Let's see if we can noodle it out.

$\frac {1,000,000 -1}7 = 142857$.

And $2\times \frac {1,000,000 -1}7 = $ what... now you point out that $100 = 7k + 2$ where $7 =14$ so $2 = 100-7k$

$2\times \frac {100,000 - 1}7= (100-7k)\frac {1,000,000-1}7 =$

$100\frac {1,000,000 - 1}7 - k(1,000,000 -1) =$

$14,285,700 - \overline{k000000} +k$.

Hmm.... that we know $k = 14$ seems to be a really nice coincidence.

$14,285,700 - 14,000,000 + 14 = 285714$ and of course the numbers are reversed.

If we can assume that if $10^{m_j} = (\text{first }m_j\text{ digits of }142857)\times 7 + j$ for $j=2,3,4,5,6$ we'd be done.

After all we'd then have

$j \times 142,857=$

$(10^{m_j} - (\text{first }m_j\text{ digits of }142857)7)\frac {1,000,000-1}7 =$

$10^{m-j}142,857 - (\text{first }m_j\text{ digits of }142857)7)1,000,000 + (\text{first }m_j\text{ digits of }142857)7)= $

$[\overline{(\text{first }m_j\text{ digits of }142857)(\text{last }6-m_j\text{ digits of }142857)\underbrace{0...0}_{m-j}}]- [\overline{(\text{first }m_j\text{ digits of }142857)000000}]+(\text{first }m_j\text{ digits of }142857)=$

$\overline{(\text{last }6-m_j\text{ digits of }142857)(\text{first }m_j\text{ digits of }142857)}$

......

So how can we show that for all $m_j \le 6 $ that $10^{m_j} = (\text{first }m_j\text{ digits of }142857)\times 7 + j$

Well.....

we know that $\frac {1,000,000 -1}7 = 142,857$.

This means $\frac 17= \frac 1{1000000}\frac {1000000}7 = \frac 1{1000000}[142,857 + \frac 17]$. That means $\frac 17 = 0.142857\overline{142,857}$.

Okay....

So if we multiply $\frac 17$ times $10^{m_j}$ we get $10^{m_j}\frac 17 = (\text{first }m_j\text{ digits of }142857) + (\text{some decimal part less than } 1)$.

So $10^{m_j} = 7 \times (\text{first }m_j\text{ digits of }142857) + 7\times(\text{some decimal part less than } 1)=$

$7 \times (\text{first }m_j\text{ digits of }142857) + (\text{some value less than } 7)$

but as $10^{m_j}$ is a natural number (not divisible by $7$) we have

$10^{m_j} = 7 \times (\text{first }m_j\text{ digits of }142857) +j$ for some $1 \le j \le 7$

And that's it.

This will actually be true of any $n$ so that $\gcd(n,10) = 1$

Take, say $\frac 1{17}= 0.\overline{0588235294117647}$

We should get $0588235294117647 \times 2..... 16$ should be the same digits shifted. Try it.

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  • $\begingroup$ $\frac{999,999}{7}=142857$. You used $1,000,000-1$ instead. But I don't understand how from $2\cdot 999,999 = 2\times \frac {1,000,000 -1}7 = 1,999,998$ we go to $285714$ via replacing $2$ with the formula for $100$. $\endgroup$
    – Jim
    Sep 22 at 19:55
  • $\begingroup$ If $100 = 7\times 14 + 2$ then $2 = 100 - 7\times 14$. So $2\times \frac {999,999}7 = (100-7\times 14)\times \frac {999999}7 = (100 - 7\times 14)\times \frac {10^6 -1}7$ Now just expand $(100-7\times 14)\times \frac{10^6 - 1}7 = 100\times \frac {10^6-1}7 - 7\times 14\times \frac {10^6-1}7= 100\times \frac {10^6-1}7 +-14\times (10^6 -1) = 100\times \frac {10^6-1}7 - 14\times 10^6 + 14$. ... Now $100\times {10^6-1}7$ will add two zeros to $142857$ to get $14285700$. And $-14\times 10^6=-14000000$ removes the leading $14$ to get $285700$ and $+14$ adds it back to the other side: $285714$. $\endgroup$
    – fleablood
    Sep 22 at 20:06
  • $\begingroup$ $1999998 = 2\cdot 999,999 \ne 2\cdot \frac {999999}7=285714$. Why did you do anything with $2\times 999999$ that was not part of the question and never part of my solution? $\endgroup$
    – fleablood
    Sep 22 at 20:08
  • $\begingroup$ It $2 = 100 - 7\times 14$ (which it does) then $$2 \times ANYTHING = (100 -7\times 14)ANYTHING = 100\times ANYTHING - 7\times 14\times ANYTHING$$. And if $$ANYTHING = 142857 = \frac {1000000 - 1}7$$ then $$2\times 142857 = 100\times 142857 - 7\times 14\times\frac {1000000 -1}7 = 100\times 142857 - 14\times 1000000 + 14$$. $$=\color{green}{14}\color{red}{2857}\color{purple}{00} - \color{green}{14}\color{red}{0000}\color{purple}{00} + \color{purple}{14}= \color{red}{2857}\color{purple}{00}$$That's all. $\endgroup$
    – fleablood
    Sep 22 at 20:11
  • $\begingroup$ Yes I got confused, sorry about that $\endgroup$
    – Jim
    Sep 23 at 20:50

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