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1/7 in decimal form is .142857

  1. Jan 12, 2005 #1


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    1/7 in decimal form is .142857
    2/7 is a rearangement of those numbers, 3/7 is too etc. anyone know why? (btw 142857 is divible by 999)
  2. jcsd
  3. Jan 12, 2005 #2
    [tex]\frac{1}{7} = 0. \dot1 4285 \dot7[/tex]

    [tex]\frac{2}{7} = 0.\dot2 8571 \dot4[/tex]

    [tex]\frac{3}{7} = 0.\dot4 2857 \dot1[/tex]

    [tex]\frac{4}{7} = 0.\dot5 7142\dot8[/tex]

    [tex]\frac{5}{7} = 0.\dot7 1428\dot5[/tex]

    [tex]\frac{6}{7} = 0.\dot8 5714\dot2[/tex]

    The pattern that I can see is that as the nominator increases the first number increases within the numbers there are. 1 then 2 then 4 then 5 etc... Also all the numbers stay in order. Why, though, I do not know.

    The Bob (2004 ©)
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3


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    Nope.1/7 is:
    [tex] \frac{1}{7}=0.(142857) [/tex] (1)
    ,which is a totally different number from the one u've written.Yours is
    [tex] 0.142757=\frac{142857}{1000000} [/tex](2)
    and has 6 decimals,while mine is
    [tex] 0.(142857)=\frac{142857}{999999}=\frac{1}{7} [/tex](3)
    and has an infinite number of decimals.

    What do u mean by rearrangement??

    PS.From (3) u get
    [tex] 142857\cdot 7=999999=999\cdot 1001=999\cdot 7\cdot 11\cdot 13 [/tex]
    ,from which u get the decomposition of 142857 in prime factors.
  5. Jan 13, 2005 #4


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    i meant rearangement of the first 6 digits, since they repeat you can think of it as 142857/999999, and incidentaly it reduces. actually i think that explains why its divisible by 999.
  6. Jan 13, 2005 #5


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    The same holds for any number [itex]n[/itex] which has the property that [itex]\frac{1}{n}[/itex] has [itex]n-1[/itex] repeating digits.
    Here's what happens:
    Code (Text):

         1 = 1 mod 7
        10 = 3 mod 7
       100 = 2 mod 7
      1000 = 6 mod 7
     10000 = 4 mod 7
    100000 = 5 mod 7
    And, calculating the digits after the decimal point is very much like multiplying the denominator by 10, and calculating the last digit and a new remainder, so you cycle through the six possible remainders.
  7. Jan 15, 2005 #6
    NateTG has it pretty well figured out, but I never did see until now. By the way, they are called cyclic numbers and can be googled.

    We have [tex]1/7 = \overline{.142857}[/tex]. (Where the overline means the period is repeated and repeated)

    The first decimal,1, represents how many times 7 goes into 10 whole , 14 represents how many times 7 goes whole into 100, 142 for 1000, etc.

    Thus if we look at 100/7 = [tex]14\overline{.285714}[/tex]

    But 100/7 = 14 +2/7, so we end up with 2/7= [tex]\overline{.285714}[/tex]

    Thus as NateTG tells us to get the full array we need to have a cycle of N-1 digits, where 10^(N-1)==1 Mod N. This will only possibly occur when N = p, a prime.

    For example, lets look at 1/21 = [tex]\overline{.047619}[/tex]

    The powers of 10^x modulo 21 are 1,10,16,13,4,19,1...for x=0,1,2,3,4,5,6...

    i.e. 10^6 == 10^0==1 Modulo 21, and the series has finished.

    Thus the cycle will repeat for some numbers, say, 16/21 = [tex]\overline{.761904}[/tex]

    But, of course, since there are only 6 possibilities, some numbers will not cycle that way, for example we have 2/21 = [tex]\overline{.095238}[/tex]
    Last edited: Jan 15, 2005
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