Take One and Divide it By 998001 For Surprising Results

Bzzt!  Skips 998.

Also, 1/9801 skips 98.

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

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

That is pretty cool.

@Kahomono and Luis: NNNNEEERRRRRRRRRRRRRD!

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

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

1/99980001=…..

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

i can see 98

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

It skips 998

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

1/9801 does skip 98 as Kahamono points out

1/81 works the same way skipping 8

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

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.

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.

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)))

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

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

You’re dividing 1 by 998001

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

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

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

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

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

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.

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

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

I don’t get it

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

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.

1/81 also does this and skips 8… interesting