If $\displaystyle x=0.1234567891011121314151617181...303132333...$

find $\displaystyle 8x$

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- Feb 29th 2008, 06:02 AM #1

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- Nov 2006
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- Feb 29th 2008, 08:18 AM #2
$\displaystyle x=0.123456789101112131415161718192021...$

This number is called Champernowne Constant.

It's given by

Since it's a transcendental number, I can't see a way to multiply it by 8..

$\displaystyle 8x = 0.987654312808897051321293745536169778594021018263 4425058667...$

- Feb 29th 2008, 02:01 PM #3
Break the number

*x*into chunks, where each chuck starts with a 0 and ends just before the next 0 (and must include some nonzero digits). Multiply each chunk by 8, and then add the chunks up.

For example, the first chunk is 0.1234567891. Multiply that by 8: $\displaystyle 0.9876543128$

The second chunk is 01112131415161718192. Multiply that by 8: $\displaystyle 08897051321293745536$

Hence the first 31 digits (30 decimal places) of 8*x*is: $\displaystyle 0.987654312808897051321293745536$

Now continue with the third chunk 022122232425262728293, fourth chunk 03132333435363738394, etc.