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Uncategorized Monday, January 23rd 2012 at 3:34 pm

Take One and Divide it By 998001 For Surprising Results

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Unfortunately, a lot of calculators are going to truncate the results. However, if you manage to get a hold of one that doesn’t, solving 1/998001 will generate all the three digit numbers from 000 to 999. And in order, no less. I have no idea how this works, but it’s a pretty neat trick and even a bit unsettling. If you’re a fan of this kind of spooky math fun, solving 1/9801 will generate all the consecutive two digit numbers. 

(Futility Closet via TYWKIWDBI)

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  • http://twitter.com/Kahomono Kahomono

    Bzzt!  Skips 998.

    Also, 1/9801 skips 98.

  • http://www.facebook.com/whydoitweet Derek Groen

    Also, you can sling in infinite pairs 9s and 0s if you like, for infinite length sequences ;).

  • Luis Ast

    You are not “solving an equation.” You are writing the fraction in decimal form.

  • Dr Coene

    That is pretty cool.

  • Jackbondnj16

    @Kahomono and Luis: NNNNEEERRRRRRRRRRRRRD!

  • Joaquin Diaz Trepat

    it’s call geekosystem… what did you expect? it’s the right place for us.. .nerds :)

  • Bob Delaney

    1/81 to over 10 places gives all numbers from 0 to 9 except 8 as a repeating decimal.

    1/9801 to over 200 places gives all numbers from 00 to 99 except 98 as a repeating decimal.

    1/998001 to over 3000 places gives all numbers from 000 to 999 except 998 as a repeating decimal.

    1/99980001 to over 4000 places gives all numbers from 0000 to 9999 except 9998 as a repeating decimal.

    1/9999800001 to over 50000 places gives all numbers from 00000 to 99999 except 99998 as a repeating decimal.

    This sequence probably continues to infinity. Can anyone prove this?

    Bob

  • York

    1/99980001=…..

  • Anonymous

    As long as no one here divides by zero, we’ll be okay.

  • http://www.facebook.com/people/Ruth-Braham/100000701832146 Ruth Braham

    i can see 98

  • Jayre

    that’s not an equation, it’s just a single number.

  • http://www.facebook.com/people/Craig-McGillivary/1570892833 Craig McGillivary

    It skips 998

  • http://www.facebook.com/people/Craig-McGillivary/1570892833 Craig McGillivary

    Here is the python code I tested it with;
    import numpy as np
    a=1000000.0
    b=998001.0
    c=[]
    for ind in np.arange(3000):
        c.append(np.floor(a/b))
        a=10*np.remainder(a,b)
    print c

  • http://www.facebook.com/people/Craig-McGillivary/1570892833 Craig McGillivary

    1/9801 does skip 98 as Kahamono points out

  • http://www.facebook.com/people/Craig-McGillivary/1570892833 Craig McGillivary

    1/81 works the same way skipping 8

  • http://www.facebook.com/people/Craig-McGillivary/1570892833 Craig McGillivary

    This isn’t a formal proof but if you treat it as a two digit number of base X then you get something like this
    1/(x*(x-2)+1)
    the first digit is 0 so this becomes
    X/(X^2-2*X+1)
    which is also zero so it becomes
    X^2/(X^2-2*X+1)
    which is
    1 rem 2x-1
    keep going
    (2X^2-x)/(x^2-2x+1)
    2 rem 3x-2
    (3x^2-2*x)/(x^2-2x+1)
    3 rem 4*x-3
    (4x^2+11*x)/(x^2-2x+1)
    4 rem 5*x-4
    5 rem 6*x-5
    6 rem 7*x-6
    7 rem 8*x-7
    8 rem 9*x-8
    .
    .
    .
    x-3 rem (x-2)*x-(x-3) -> x-3 rem x^2-3*x+3
    x-2 rem (x-1)*x-(x-2) -> x-2 rem x^2-2*x+2 -> x-1 rem 1
    0 rem X
    1 rem 2x-1

    So basically it works for all bases of X

  • Notanemail

    One way to find such a number is to use the formula for a geometric series: {sum from k=0 to n of (r^k)} = (r^(n+1)-1)/(r-1).  Differentiating both sides and then multiplying by r gives a formula, f(r,n)= {sum from k= 0 to n of k*r^k}.  setting r = 1/1000, its then easy to set a computer program looking for integer values of 1/f(1/1000,n).  998001 is one such value. This would work for any longer sequence of numbers.  you would just have to adjust r and set your computer looking for integer values of 1/f(r,n).  Thankfully some of them are relatively small.

  • Notanemail

    actually, thats silly. why not just look at the limit of the sequence 1/f(1/1000,n) as n goes to infinity.  This gives 998001. letting n go to infinity in the formula f(r,n) defined above also allows you to derive the sequence bob delaney mentions because:

    999^2=998001

    9999^2=99980001

    99999^2=9999800001 etc.

  • Adam

    It would seem (from a few examples) that all consecutive n-digit numbers will appear in the decimal expansion of 1/9..980..01, where there are n-1 9s and n-1 0 (for example, 1/81 expands to 0.0123456789…

    Here’s some python code to calculate the number n such that the decimal expansion of 1/n contains all the d-digit numbers in order:

    def series(digits): fmt = ‘{:0′+str(digits)+’}’ return int(’1′+(’0′*(10**digits*digits)))/int(”.join(fmt.format(i) for i in range(10**digits)))

  • Adam

    The python code got messed up, not sure how to fix it. Should contain three lines.

  • guest1

    thats cool. but how do you know that it doesn’t work? can you verify this?

  • http://profiles.google.com/flyingleech Andrew T

    You’re dividing 1 by 998001

  • Anonymous

    sum( from n=1 to inf) ( n*10^^(-3(n+1))

    =1 /(1-10^^(-3))^^2 = 1/(999)^^2 = 1/998001

  • Robert Delaney

    Looks good! And I suspect that form generalizes as 3 is replaced by 4, 5, …

  • Pseudonym Anonymous

    Maybe something to do with the fact that 998001=999^2? Or that 9801=99^2?

  • Thecozze

    is it not 7[9, 8]0, 81 … that you see?

  • theran davis

    any time an = sign is involved it is an equation so 1/99801=x and your trying to find x. so it is an equation.

  • Robert Delaney

    Alan Listoe has shown why in the decimal expansion of 1/998001 the pattern 998 does not appear. 

    In the series:

    sum( from n=1 to inf) ( n*10^^(-3(n+1)) 

    we have up to and including n = 999, where I’ve added spaces for clarity:

    … 996 997 998 999

    so the next term for n=1000 will add 1000 so that 000 is added after the 999, but the 1 is added to the 999, which gives 000 and a carry of 1 which is added ti the 998 making it 999, so now we have:

    … 996 997 999 000 000

    then the n=1001 term adds so that the sum is now:

    … 996 997 999 000 001 001

    and the next term n=1002 gives:

    … 996 997 999 000 001 002 002

    and so on. Not the same, but something similar, must happen when n=2000 is reached since we have a 2 being added in instead of a 1. I haven’t examined that closely as the series continues.

    Bob

  • Zaciscool576

    Technically it is an equation. You are dividing 1 by 998001.

  • Lucas Steffen

    I don’t get it

  • Rover12421

    1/99980001
    1/9999800001
    1/999998000001
    …..
    1/(9…)n8(0…)n1
    ….

  • Nah

    You kind of ARE solving an equation.  1 divided by 998,001.  It’s essentially the same thing as writing it in decimal form.  Just different ways of looking at it.

  • Jimxfitz

    1/81 also does this and skips 8… interesting

  • Steve

    Nope, it’s definitely 98.

    its down from the top by 5 lines
    097098099100 etc. 

  • Lucy

    If you notice that 998001 is 999 squared, you can see the reason why.

    1/998001 = 0.000001 x 1/(1-0.001)^2.

    Then if you use the power series expansion 1/(1-x) = 1+x+x^2+x^3+… this means
    1/998001 = 0.000001 x (1+0.001+0.001^2+….) x (1+0.001+0.001^2+…)
    and so you can count the number of ways to get a certain power of 0.001 in this product…

  • http://www.facebook.com/profile.php?id=100000102801320 Keandre Espina

    this doesn’t “generate all the three digit numbers from 000 to 999″ it skips 998 still cool though

  • Anthony McWilliams

     It skips 998, not 98. Numberphile showed why on their YouTube channel – search for “numberphile 998001″.

  • http://twitter.com/kbutti はぶ

    If you using Mac, then type this on Terminal:

    ruby -r bigdecimal -e “puts BigDecimal(’1′).div(998001, 3000)”

  • http://www.onlineroulettebetting.com/ Onlineroulettebetting

    this is amazing I love number puzzles like this, do people memorize this number like they memorize pi?

  • cheap bras

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  • Rickant1

    Sorry to bring the math into this but this is actually not so unusual and is the result of 1 being divided by the square of 999 or 99 and works with other numbers that consist of only 9s. It is part of a mathematic characteristic that allowed any desired number to “become” it’s own repeating irrational decimal. One example would be 345/999. The answer would be 0.345345345345… There is a video on YouTube by numberphile that explains it much better. It is very interesting but not really a big deal.

  • Mr. Briggs

    For more mindfuckery, try dividing 1 by 998999. Or 999998.

  • Rance Mohanitz

    Repeating what Kahamono said: 1/9801 skips 98. 1/998001 is the one that skips 998.

  • Littlet102030

    Pi does it too.

  • greg

    he says it skips 98 for a different fraction

  • Alex Evans

    (1/9)^2 skips 8
    (1/99)^2 skips 98
    (1/999)^2 skips 998
    (1/9999^2) skips 9998
     

  • Jeff

    I prefer 1 / (9 * 9) or 1/81 which equals 0.1234567890123456789…

  • Tucker

    not an equation, it’s an expression. an equation has an equals sign.

  • Hugo

    Also, if you try to divide by:
    999991, you get the powers of 9
    999992 – powers of 8
    999993 – powers of 7
    and so on…
    It’s limited to 6 digits. To add more digits, just add more nines at the begin.

  • Herp Derp

    Actually 1/81 is 0.12345679012345679…. It actually skips the 8.

  • Aaron Li

    One three digit number is not in it: 998

  • Guest

    Here’s an equation for you. You – Nerds = No computer fro you to sit in front of, Face-booking your miserable existence away. :P

  • Anonymous

    huh?

  • Anonymous

    lol

  • Anonymous

    You math weenies are scaring me…..

  • Anonymous

    stop the madness!

  • josie

    I am so showing this to my math teacher!

  • Man Friday

    Is that what remains after feds have their hands on my pay?

  • David J

    it’s actually not that surprising cause any rational number can be expressed as a fraction, by definition