# 1/998001

#### Jameson

Staff member
$$\displaystyle \frac{1}{998001}$$ has some interesting properties. It will list out every 3 digit number except for 998.

It has the form 0.001002003004...

Here's a video on some more details if you're interested.

#### MarkFL

Staff member
soroban once posted this observation and asked:

This is just one of a family of such fractions.
Can you determine the underlying characteristic?
I looked at the factorization of the denominator, and found:

$\displaystyle 998001=999^2$

So, next I looked at:

$\displaystyle\frac{012345679}{999999999}=\frac{1}{9^2}$

Thus, I conjecture that the family you speak of is:

$\displaystyle\frac{1}{(10^n-1)^2}$ where $\displaystyle n\in\mathbb N$

where the decimal representation contains all of the $n$ digit numbers except $\displaystyle 10^n-2$ and the period is $\displaystyle n(10^n-1)$.

#### Jameson

Staff member
I had a feeling when I was posting that soroban might have already said something about this Yes, you are correct about this form. For example $$\displaystyle \frac{1}{9999^2}$$ gives all of the 4 digit numbers.

In the video I posted in the OP they discussed an easy way to write recurring decimals as fractions that I'm sure we are all familiar with this method but it's still worth writing for those who aren't.

If you want to write some repeating decimal all you do is write the string as the numerator to a fraction and then in the denominator write the same number of 9's. For example, if you want to write $$\displaystyle .\overline{12436298}$$ as a fraction. It's simply $$\displaystyle \frac{12436298}{99999999}$$. Of course this method can be derived quite easily. If $$\displaystyle x=.12436298$$ then $100000000x=12436298.\overline{12436298}$ so $9999999x=12436298$ and we can solve for x directly. However it's nice to know a shortcut to not have to calculate it "the long way" everytime.

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#### Bacterius

##### Well-known member
MHB Math Helper
This seems to generalize to an arbitrary base $b$, where:

$$\frac{1}{(b^n - 1)^2}$$

has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.

But I only checked a couple of bases, so this could be wrong.

#### MarkFL

Staff member
This seems to generalize to an arbitrary base $b$, where:

$$\frac{1}{(b^n - 1)^2}$$

has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.

But I only checked a couple of bases, so this could be wrong.
I believe you are right, as the method Jameson outlined above can be generalized to any valid radix.

#### alane1994

##### Active member
Not much to contribute to this conversation, I was just popping in to say that this is really interesting!