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Strange number pattern

  1. May 16, 2003 #1
    1/7 = 0.142857142857....
    2/7 = 0.285714285714...
    3/7 = 0.428571428571...
    4/7 = 0.571428571428...
    5/7 = 0.714285714285....
    6/7 = 0.857142857142...

    That's strange.
     
    Last edited: May 16, 2003
  2. jcsd
  3. May 16, 2003 #2
    why...

    is that?
     
  4. May 16, 2003 #3
    All fractions repeat or terminate eventually.

    Here, we know it will repeat after n digits if 10^n / 7 = whole number + 1/7.

    ie, 10^n = 1 (mod 7)
    true for n=6

    10^6/7 = 142857 + 1/7
     
  5. May 16, 2003 #4
    why strange?
     
  6. May 16, 2003 #5
    Does this also mean that PI will terminate sooner or later?
     
  7. May 16, 2003 #6

    selfAdjoint

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    By fractions he meant one whole number divided by another one. There are numbers (irrational numbers) that can't be gotten that way. Square root of 2 and pi are a couple of well known ones. Their decimal expansions are non-repeating.
     
  8. May 16, 2003 #7

    Integral

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    All RATIONAL numbers can be represented as a repeating or finite length decimal. A rational number is one which can be represented as a ratio of 2 integers. Thus all fractions and integers.

    Pi OTH is not a rational number, it is a irrational, so cannot be represented as a finite length or repeating decimal.

    The Real number system is composed of both the rational and irrational numbers.
     
  9. May 16, 2003 #8
    This is an interesting

    but well studied observation. All the numbers that divide into 99 repeat after a period of at most two, all that divide into 999 by at most three, ect., so that means that seven divides into 999999. Other numbers systems also work this way, nothing peculiar about base ten. If you have a sequence of 1's that is a prime number, then multiply it by 9, then it's only factors are the prime× 3×3, and the prime will divide with the period of the sequence length.
     
  10. May 16, 2003 #9
    Fermat's little theorem. :wink:
    So why's there rotation of the digits when a/7 (a isn't 7's multiple)

    |[rr]| < |Q| (where Q stands for set of rational numbers)
    I've heard that pi, e, golden ration, etc are trancedental(sp?) numbers, but the proof is far beyond my level.
     
  11. May 16, 2003 #10
    as you can see, the numbers only contain "142857" and they are rotated in a periodic manner. (I've edited my first post a bit)
     
  12. May 16, 2003 #11

    Hurkyl

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    Try performing the long division necessary to get the digits.

    One you might notice is that there are only a finite set of possible remainders, and any particular remainder is always followed by the same next remainder. It's easy to see why; at each step you add a 0 and divide again, so same remainder means same next step.

    For example, I'll do it with 7 (with somewhat modified notation; I hope it's clear):

    7 | 1 -> 0 R 1
    7 | 10 -> 0.1 R 3
    7 | 30 -> 0.14 R 2
    7 | 20 -> 0.142 R 6
    7 | 60 -> 0.1428 R 4
    7 | 40 -> 0.14285 R 5
    7 | 50 -> 0.142857 R 1
    7 | 10 -> 0.1428571 R 3
    ...

    The sequence of remainders will repeat forever (1, 3, 2, 6, 4, 5), which will forever generate the sequence of digits (142857).

    There are only 7 possible remainders when you divide by 7; 0, 1, 2, 3, 4, 5, and 6. The last 6 of them are in the above cycle, so any time you get one of those remainders, the following digits will follow that cycle. The only other possibility is that 7 evenly divides the number in question (such as 14 / 7), where there is only one remainder in the cycle (0) and the sequence of digits is simply an infinite string of 0's.


    Every other number behaves similarly, except 7 is somewhat special by having one large cycle of non zeroes. For example, with the number 4:

    Starting with a remainder of 0, 00 / 4 = 0 R 0, so remainder 0 progresses to remainder 0 yielding digit 0

    Starting with a remainder of 1, 10 / 4 = 2 R 2, so remainder 1 progresses to remainder 2 yielding digit 2

    Starting with a remainder of 2, 20 / 4 = 5 R 0, so remainder 2 progresses to remainder 0 yielding digit 5

    Starting with a remainder of 3, 30 / 4 = 7 R 2, so remainer 3 progresses to remainder 2 yielding digit 7

    So, if you were computing something like 3/4, you get a remainder of 3, which yields remainder 2 then an infinite sequence of 0's, which gives the digits 7, 5, 0, 0, 0...

    i.e. 3 / 4 = 0.75000000...


    Other numbers have different cycle structures. 6 has several cycles, which can be observed in its somewhat interesting diversity of expansions:
    0/6 = 0.00000...
    1/6 = 0.16666...
    2/6 = 0.33333...
    3/6 = 0.50000...
    4/6 = 0.66666...
    5/6 = 0.83333...


    The number 11 has a whole buch of two cycles, yielding the familiar expansions:

    1/11 = 0.090909...
    2/11 = 0.181818...
    3/11 = 0.272727...
    et cetera


    The number 13 is more similar to 7; it yields two seperate 6-cycles, but not a 12 cycle (of course, it also has the trivial all 0's cycle). The two corresponding patterns are:

    (076923) and (153846)

    And any decimal expansion of a fraction with denominator 13 will end in a repeating sequence of one of those cycles of digits. For example:

    4/13 = 0.3(076923)07692...


    I don't know if any larger numbers have a complete cycle like 7 does; it has been a long time since I played with this, and I don't think I went past 13.
     
    Last edited: May 16, 2003
  13. May 16, 2003 #12
    Thanks Hurkyl, that's interesting.

    well in fact I've come across a question from a math contest(group section), which initiated me to start this post.

    Question :
    Given that abcdef is a six digit number(each letter represents one digit) and all of them are distinct and non of them equals 0. Find abcdef. As a remark, abcdef can be uniquely determined.
    bcdefa=2*abcdef
    cdefab=3*abcdef
    defabc=4*abcdef
    efabcd=5*abcdef
    fabcde=6*abcdef

    My friend knew the special number 142857 and wrote it down immediately, but I'm thinking what if we don't know 142857, how can we tackle this problem?

    By the way, hmm... for example the fraction 5/99783, as it is a rational number, there must be a repeating cycle, right? Can we determine how many digits later will the repeating cycle emerge? I think I've heard that(long long long time ago and I'm not sure whether this memory of mine's valid!) we can proof that the number of digits before the repeating part of a rational fraction emerges doesn't exceed N, where N is a very large number. (I hope you got what I mean!!) Has anyone heard of it also ?
     
  14. May 17, 2003 #13

    Hurkyl

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    Hrm, you're misstated the problem; the problem you stated has no solution.

    proof:

    Because efabcd=5*abcdef, d must either be 5 or 0. Since d isn't allowed to be 0, d is 5, and f must be odd.

    defabc=4*abcdef and fabcde=6*abcdef imply that
    fabcde = (3/2)*defabc < (3/2)*599999 < 900000
    so f < 9

    However, fabcde = (3/2)*defabc implies that f >= d, so f = 7.

    Now that we know f = 7, we can find all the other digits:
    a = 2 * f mod 10 = 4
    b = 1
    c = 8
    d = 5
    e = 2

    so the number (if it exists) must be 418527, but you can easily check it doesn't work. (in particular, you can easily show from the problem you stated that a<b<c<d<e<f)


    N, in fact, is the number we're dividing by!

    When you divide something by x, there are only x possible remainders, so you cannot possibly have more than x digits before the first digit of a repetition.
     
  15. May 17, 2003 #14
    I dont see how its possible for the letters to equal anything

    Say:

    a = 4
    b = 1
    c = 8
    d = 5
    e = 2
    f = 7


    Now the problem is:

    bcdefa=2*abcdef
    cdefab=3*abcdef
    defabc=4*abcdef
    efabcd=5*abcdef
    fabcde=6*abcdef


    now if we put in the variables it would be:

    (1*8*5*2*7*4)=2*(4*1*8*5*2*7) =
    2240=4480
    ?

    Thats not right, is it?
     
  16. May 17, 2003 #15

    Hurkyl

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    For the problem (as described), a string of variables is supposed to mean that each variable represents a digit of the number, not that you multiply the variables.
     
  17. May 17, 2003 #16
    modified question

    Modified question:
    cdefab=2*abcdef
    bcdefa=3*abcdef
    efabcd=4*abcdef
    fabcde=5*abcdef
    defabc=6*abcdef

    Thanks Hurkyl, clear explanation!

    I'm drilling a hole.....
     
  18. May 17, 2003 #17
    Yes, that is why

    999999/7 = 142857
     
  19. May 18, 2003 #18
    gotcha, thanks.
     
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