What are examples of integers that halve their values when all 9's in their decimal representations are replaced with 0's?
Not the usual decimal representation, but...
$1.9999999....$ is double $1.00000... $
I originally did not point this out as to not complicate the answer, but we can make as many examples as we want:
Just move the decimal point in the previous answer to the right:
$19.999999....$ is double $10.0000... $
$199.99999....$ is double $100.000... $
and so on.
One solution is:
0... and this is also the only whole-number solution!
* If you take a number containing at least one 9 and subtract the same number with all 9s replaced by 0s, then the difference between them will consist of some sequence of 0s and 9s.
* Specifically, the difference contains a 9 in every place value in which a 9 in the minuend (the original number, before the minus sign) is replaced by a 0 in the subtrahend (the replaced number, after the minus sign). All other digits of the difference are 0s.
* So the difference has a 9 in each place value where the minuend has a 9 and the subtrahend has a 0, and the difference has 0s in all other place values.
* If the substitution process halves the minuend (the original number), then the difference must be equal to the subtrahend (the subtracted and replaced number, after the minus sign).
* If the subtrahend and the difference are equal, then all their digits -- in all place values -- must be identical.
* But recall that wherever the difference has a 9, the subtrahend has a 0.
* So the difference can contain only 9s and 0s, and if the minuend is halved, it can't contain any 9s. Therefore, it must contain only 0s, making it equal to 0.
* Therefore, the subtrahend is also 0.
* And solving x - 0 = 0 for x, the minuend is also 0.