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Math Help - Repetends and families

  1. #1
    Junior Member Bartimaeus's Avatar
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    Question Repetends and families

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

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

    Thanks
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  2. #2
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    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.

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

    We have: . \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.

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


    There are nine families:

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

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

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

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

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

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

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

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

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

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    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.

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  3. #3
    Junior Member Bartimaeus's Avatar
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    Thumbs up

    Shortcuts?
    Yes, i am interested in your findings, thanks.
    That would help me out a great deal.
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  4. #4
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    Hello again, Bartimaeus!


    Here is the main shortcut I used . . .

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

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

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

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


    Move the decimal point again: 0.36986301

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

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


    Move the decimal point again: 0.69863913

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

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


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



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

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

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

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


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

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


    So you can see how this speeded up the process.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    A few more tips on repeating decimals. . .

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

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

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


    Then how come: \frac{1}{11}\;=\;0.\overline{09} and \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: 99,\;999,\;9999,\;99999,\;\hdots

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


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

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

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


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

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

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

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  5. #5
    Junior Member Bartimaeus's Avatar
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    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?
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  6. #6
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    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 13.

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

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

    Hence, denominator 13 has two families.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Even more trivia . . .

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


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

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

    . . \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.

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

    . . . \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 999...)

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

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

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