# Thread: Repetends and families

1. ## Repetends and families

List the families of repetends of fractions with the denominator 73. Give an example of a fraction which has each repetend.

- The correct number of families
- Any six correct repetends
- The remaining correct repetends
- A correct member of each of the families

Thanks

2. Hello, Bartimaeus!

I'm not sure of the terminology, but I think I understand . . .

List the families of repetends of fractions with the denominator 73.
Give an example of a fraction which has each repetend.

- The correct number of families
- Any six correct repetends
- The remaining correct repetends
- A correct member of each of the families

We have: .$\displaystyle \frac{1}{73}\,=\,0.\overline{01369863}$ ... a repeating decimal with an 8-digit cycle.

If we "cycle" the digits, we get seven more fractions of the family.

. . $\displaystyle \begin{array}{ccccccc}0.13698630 = \frac{10}{73}\\ \qquad 0.36986301 = \frac{23}{73}\\ \qquad\qquad 0.69863013 = \frac{51}{73}\\ \qquad\qquad\qquad 0.98630136 = \frac{72}{73}\\ \qquad\qquad\qquad\qquad 0.86301369 = \frac{63}{73}\\ \qquad\qquad\qquad\qquad\qquad 0.63013698 = \frac{46}{73}\\ \qquad\qquad\qquad\qquad\qquad\qquad 0.30136986 = \frac{22}{73}\end{array}$

I will abbreviate this: .$\displaystyle \frac{1}{73}= 0.01369863\quad\{10,23,51,72,63,46,22\}$

There are nine families:

$\displaystyle \frac{1}{73} = 0.01369863\quad \{10,23,51,72,63,46,22\}$

$\displaystyle \frac{2}{73} = 0.02739726\quad \{20,54,29,71,53,19,44\}$

$\displaystyle \frac{3}{73} = 0.04109589\quad \{30,8,7,70,43,65,66\}$

$\displaystyle \frac{4}{73}= 0.05479452\quad \{40,35,58,69,33,38,15\}$

$\displaystyle \frac{5}{73} = 0.06849315\quad \{50,62,36,68,23,11,37\}$

$\displaystyle \frac{6}{73} = 0.08219178\quad \{60,16,14,67,13,57,59\}$

$\displaystyle \frac{9}{73} = 0.12328767\quad \{17,24,21,64,56,49,52\}$

$\displaystyle \frac{12}{73} = 0.16438356\quad \{47,32,28,61,26,41,45\}$

$\displaystyle \frac{18}{73} = 0.24657534\quad \{34,48,42,55,39,25,30\}$

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Note: I did not crank out and examine all seventy-two decimals.
. . . . .I "invented" some labor-saving shortcuts.
If you are interested in these methods, please let me know.

3. Shortcuts?
Yes, i am interested in your findings, thanks.
That would help me out a great deal.

4. Hello again, Bartimaeus!

Here is the main shortcut I used . . .

First I found: $\displaystyle \frac{1}{73} = 0.\overline{01369863}$

I "moved" the decimal one place to the right (cycling the digits): $\displaystyle 0.13698630$

. . Multiply by 73: .$\displaystyle 0.13698630 \times 73\:=\:9.9999999 \:\approx\;10$

. . Hence: .$\displaystyle 0.\overline{13698630}\:=\:\frac{10}{73}$

Move the decimal point again: $\displaystyle 0.36986301$

. . Multiply by 73: .$\displaystyle 0.36986301 \times 73 \:=\:26.99999972\:=\:\approx 27$

. . Hence: .$\displaystyle 0.\overline{36986391} \:=\:\frac{27}{73}$

Move the decimal point again: $\displaystyle 0.69863913$

. . Multiply by 73: .$\displaystyle 0.69863913 \times 73 \:=\:50.00000040\:\approx\:51$

. . Hence: .$\displaystyle 0.\overline69863013} \:=\:\frac{51}{73}$

Continuing in this manner, I found that this first "family" had these numerators:
. . . . . $\displaystyle 1,\;10,\;23,\;51,\;72,\;63,\;46,\;22$
(I kept a list of these numerators off on the side.)

Then I got: .$\displaystyle \frac{2}{73} = 0.\overline{02739726}$

Move the decimal point and multiply by 73: .$\displaystyle 0.27397260 \times 73 \:=\:19.9999998\:\approx\:20$
. . Hence: .$\displaystyle 0.\overline{27397260} \:= \:\frac{20}{73}$

Move the decimal point and multiply by 73: .$\displaystyle 0.73972602 \times 73 \:=\:53.99999946\:\approx\:54$
. . Hence: .$\displaystyle 0.\overline{73972602} \:= \:\frac{54}{73}$

And found that this family had numerators: .$\displaystyle \{2,\;20,\;54,\,29,\;71,\;53,\;19,\;44\}$

I continued with $\displaystyle \frac{3}{73},\;\frac{4}{73},\;\frac{5}{73},\; \frac{6}{73}$

However, $\displaystyle \frac{7}{73}$ and $\displaystyle \frac{8}{73}$ are already in the family of $\displaystyle \frac{3}{73}$
. . (That's why I kept a list of numerators, you see.)

So you can see how this speeded up the process.

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A few more tips on repeating decimals. . .

If $\displaystyle p$ is a prime, then $\displaystyle \frac{1}{p}$ can have at most a $\displaystyle (p\!-\!1)$-digit cycle.

For example: .$\displaystyle \frac{1}{7} \;= \;0.\overline{142857}$

. . . . . And: .$\displaystyle \frac{1}{17} \;= \;0.\overline{0588235294117647}$

Then how come: $\displaystyle \frac{1}{11}\;=\;0.\overline{09}$ and $\displaystyle \frac{1}{13} \;=\;0.\overline{076923}$ ?

There is a theorem that covers this . . . Here's my baby-talk version of it.

Divide the prime into a "string of 9's" . . . that is: $\displaystyle 99,\;999,\;9999,\;99999,\;\hdots$

When the division "comes out even", the number of 9's is the length of the cycle.

So: .$\displaystyle (9999\hdots9) \div 17$ requires sixteen 9's to come out exact.
. . Hence: .$\displaystyle \frac{1}{17}$ has a 16-digit cycle.

However: .$\displaystyle 999999 \div 13$ comes out even.
. . Hence, $\displaystyle \frac{1}{13}$ has a six-digit cycle.

And: .$\displaystyle 999 \div 37$ comes out even.
. . Hence, $\displaystyle \frac{1}{37}$ has a three-digit cycle.

This explains why $\displaystyle \frac{1}{73}$ does not have a 72-digit cycle.
. . If it did, there would be one "family".

It turns out that $\displaystyle 99,999,999 \div 73$ comes out even.
. . Hence, $\displaystyle \frac{1}{73}$ has an eight-digit cycle.

That's how I knew that there would be nine families.

5. G'day again Soroban,
I'm quite sure, but want to double check with you about the families.
A family is where the repetends have the same digits in a cycling order, correct?

6. Hello, Bartimaeus!

A family is where the repetends have the same digits in a cycling order, correct?

As I said, I'm not sure of the terminology used by your course/book/teacher,
. . but that is my interpretation of "family".

For example: denominator $\displaystyle 13.$

$\displaystyle \frac{1}{13}\,=\,0.\overline{076923}$
. . If we "cycle" the decimal, the numerators are: $\displaystyle \{1,\;10,\;9,\;12,\;3,\;4\}$

$\displaystyle \frac{2}{13} \,=\,0.\overline{153846}$
. . These numerators are: $\displaystyle \{2,\;7,\;5,\;11,\;6,\;8\}$

Hence, denominator $\displaystyle 13$ has two families.

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Even more trivia . . .

For prime $\displaystyle p$, the fraction $\displaystyle \frac{1}{p}$ will have at most $\displaystyle p\!-\!1$ digits in its cycle.

(1) If the cycle is shorter, say, length $\displaystyle m$, then $\displaystyle m$ is a divisor of $\displaystyle p\!-\!1.$

. . $\displaystyle \frac{1}{73}$ could have had a 72-digit cycle, but it has an 8-digit cycle.

. . $\displaystyle \frac{1}{13}$ could have had a 12-digit cycle, but it has a 6-digit cycle.

(2) The sum of the digits of a cycle is a multiple of 9.

. . .$\displaystyle \frac{1}{7} = 0.\overline{142857}\quad\Rightarrow\quad 1+4+2+8+5+7 = 27$

. . . $\displaystyle \frac{1}{41} = 0.\overline{02439}\quad\Rightarrow\quad0+2+4+3+9 = 18$

(3) If the cycle has an even number of digits,
. . .the first half and the second half are "9-conjugates".
. . .(Their sum is of the form $\displaystyle 999...$)

. . .$\displaystyle \frac{1}{13} = 0.\overline{076\:923}\quad\Rightarrow\quad 076 + 923 = 999$

. . .$\displaystyle \frac{1}{137} = 0.\overline{0072\:9927}\quad\Rightarrow\quad 0072 + 9927 = 9999$