What value do you get when you convert \(\frac {1}{81}\)to decimal? You get \(0.0123456790123456790...\).

What value do you get when you convert \(\frac {1}{9801}\) to decimal? You get \(0.000102030405060708091011...9697990001\).

What value do you get when you convert \(\frac {1}{998001}\) to decimal? You get \(0.000001002003...100101102...996997999000...\).

These decimals list every \(n\) digit numbers (81 is 1, 9801 is 2, 998001 is 3, etc.) apart from the second last number. There is a pattern to find one of these fractions.

Can you see something special about the denominators? \(81\) is \(9^2\), \(9801\) is \(99^2\), \(998001\) is \(999^2\).

This means that if you did \(\frac {1}{99980001}\) you would get \(0.0000000100020003...9996999799990000...\).

Can you find a fraction that, when converted to decimal, lists every \(n\) digit number?

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TopNewestWatch the numberphile video which explains this very well at :http://www.youtube.com/watch?v=daro6K6mym8

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This is a result of generating functions, and plugging \(x=\frac{1}{10}\) into the function. For example, the generating function for the Fibonacci sequence is \(\frac{x}{1-x-x^2}\). If we plug in \(x=\frac{1}{10}\), we get \(\frac{1/10}{1-(1/10)-(1/10)^2}=\frac{10}{100-10-1}=\frac{10}{89}=0.11235\dots\).

To answer your question, sure you can. If you want to list every \(n\)-digit number, you'll want to have the sum \(\sum_{i=1}^\infty i\times10^{-in-n}\). Recall that the generating function for \(i\) is \(\frac{1}{(1-x)^2}\), so we'll have \(\frac{10^{-n}}{(1-10^{-n})^2}\).

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Though, for the Fibonacci sequence, note that with \( x = \frac{1}{10} \), you 'add' the tens digit to the preceding units digit, so you don't get the sequence of \(0.112358132134\ldots \), but instead \( \frac{10}{89} = 0.1123595\ldots \). Ah, if only patterns were verified by checking the first 5 terms.

Here's a spinoff question.

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Irrational. I would provide a proof,but then I would be guilty of stealing your answer from MSE.

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You see the effects of carrying the digit, yes, but it seemed more magical to post the first 5 digits.

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