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Here is what we have:

 1/7 = 0.142857...
1/11 = 0.090909...
1/13 = 0.076923...

Notice that if you add the first three digits to the next three digits, you always get 999:

142 + 857 = 999
090 + 909 = 999
076 + 923 = 999

Oddly, the same thing happens with 2/7, 2/11, 2/13, and so on, as the numerator increases: adding the first three digits to the next three digit results in 999.

OK. Multiplying the denominators 7 x 11 x 13 results in 1001.

What is the relationship between the fraction series x/7, x/11, x/13, and the result of multiplying the denominators?

(This is taken from an example in the book "Ten Ways to Destroy the Imagination of Your Child" by Anthony Esolen. The examples above are presented separately, and the reader is encouraged to be imaginative in discovering how they are related. After almost 20 years of playing math/CS games, I suppose I still have no imagination to solve this kind of thing.)

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    $\begingroup$ Hint: $\;1001 \cdot 1/7 = (1 + 1000)\cdot 1/7 = 0.\overline{142 857} + 142.\overline{857 142} = \color{red}{142.\overline{142 857 + 857 142}}\,$. On the other hand $1000 \cdot 1/7 = 143 = \color{red}{142.\overline{9}}\,$. $\endgroup$ – dxiv Sep 1 '18 at 2:53
  • $\begingroup$ I propose to call it $\underline{Scheherazade \text{ }Phenomenon}$... $\endgroup$ – DVD Sep 9 '18 at 21:45
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The relationship is that $\frac17=\frac{11\times13}{1001}$, $\frac1{11}=\frac{7\times13}{1001}$ and $\frac1{13}=\frac{7\times11}{1001}$, and the pattern of adding three digits to the next three digits to get $999$ works for any fraction $\frac{n}{1001}$ (apart from the ones that are integers, like $\frac{1001}{1001}$ or $\frac{2002}{1001}$) and for no other number.

Let's see what multiplying any number by $1001$ does. We have $$ 1001x=(1000+1)x=1000x+x $$ So multiplying a number $x$ by $1001$ is the same as taking two copies of $x$, multiply one of them by $1000$ (which has the effect of moving all the digits of $x$ three places to the left), then add them together.

Now, what's special about the numbers $\frac n{1001}$ (again, apart from the ones which are already integers) is that they are the only numbers which are not integers, but multiplying them by $1001$ makes them integers. If we use the above interpretation of multiplying by $1001$, the only way that can happen is if after doing the addition, we're left with something with decimal part $.999999\ldots$ (you can try to find examples which make it $.000000\ldots$, but you will not succeed). This means exactly that any three digits of the decimal expansion of $\frac n{1001}$, plus the next three digits must add up to $999$.

(They could, conceivably, add up to $1998$, and let the $1$ carry into the next set of three, but since $1998=999+999$, that means that we would have to start with $.999999\ldots$ already, meaning what we have is an integer. And we're not considering those.)

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  • $\begingroup$ A corresponding pattern holds true for any denominator of the form $10^k+1$ with $k\geq1$: for $\frac n{11}$, one digit plus the next is $9$. For $\frac n{101}$, two digits plus the next two gives $99$. And so on. You may also find it intriguing to study the expansions of fractions with denominators $10^k-1$. $\endgroup$ – Arthur Sep 1 '18 at 4:25
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Knowing how this phenomenon came to happen (e.g. as explained in Arthurs answer), you can construct similar ones. E.g. You can factor

$$10^\color{red}4+1=73\times 137.$$

You will observe that subdividing the decimal digits of the reciprocals of these factors into blocks of length $\color{red}{\text{four}}$ will give you sums of $9,\!999$:

\begin{align} 1/73& = 0.\underline{0136}\overline{9863}\underline{0136}\overline{9863}...\\[1ex] 1/137 &= 0.\underline{0072}\overline{9927}\underline{0072}\overline{9927}... \end{align}


Further, $1/7$, $1/11$ and $1/13$ also are special for other block lengths: e.g. they give the sum $999,\!999,\!999$ when using block length $\color{blue}{\text{nine}}$ (not very surprising). However, there are other such numbers: $1/19$ and $1/52579$. The reason is that

$$10^\color{blue}9+1=7\times 11\times 13\times 19\times 52579.$$

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  • $\begingroup$ Oh wow. This compliments Arthur's answer so well! While his answer makes sense to me, I have had trouble applying his explanation to the general situation, so that I could construct (or discover) similar patterns in other number sets (and until I can do that, I can't really say that I understand the phenomenon). Thank you! $\endgroup$ – JonathanDavidArndt Sep 5 '18 at 11:42
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(I). A number $R$ between $0$ and $1$ is represented decimally as $0.ABCDEFABCDEF...$ iff $10^6R$ is represented as $ABCDEF. ABCDEFABCDEF...$ Subtracting, we see that $(10^6-1)R$ is represented as $ABCDEF.$

(II). The digit-sequence $ABC$ represents the number $X=10^2A+10B +C$ and the digit-sequence $DEF$ represents the number $Y=10^2D+10E+F .$ Now $Y=999-X$ iff the digit-sequence $ABCDEF$ represents $(1000 X)+(999-X)=(999)(1+X). $

(III). Combining (I) and (II) we see that $X+Y=999$ iff $ (10^6-1)R=(999)(1+X)$ iff $(999)(1001)R=(999)(1+X)$ iff $$R=\frac {1+X}{1001}=\frac {1+X}{(7)(11)(13)}.$$

Now $1+X=1+(10^2A+10B+C)$ can be any integer from $1$ to $1000.$ For example if $X=285$ then $(1+X)/1001=286/1001=2/7.$ Or if $X=64$ then $(1+X)/1001=65/1001=5/77.$

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