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- #1

- Jan 26, 2012

- 4,043

It has the form 0.001002003004...

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

[video=youtube;daro6K6mym8]998001[/video]

- Thread starter Jameson
- Start date

- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,043

It has the form 0.001002003004...

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

[video=youtube;daro6K6mym8]998001[/video]

- Admin
- #2

I looked at the factorization of the denominator, and found:This is just one of a family of such fractions.

Can you determine the underlying characteristic?

$\displaystyle 998001=999^2$

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 $\displaystyle n(10^n-1)$.

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- #3

- Jan 26, 2012

- 4,043

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.

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.

Last edited:

- Jan 26, 2012

- 644

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

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.

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- #5

I believe you are right, as the method

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

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.

- Oct 16, 2012

- 126