Clockwork Powers

Here’s another problem I’ve been looking at lately:

Find a number $x$, such that if $j$ is any digit from 0 to 9, $x^{j}$ has $j$ as its first digit past the decimal point. For instance, 63.18 works for the first power $(63.18^{\b{1}} = 63.\b{1}8)$ and the third power $(63.18^{\b{3}} = 252196.\b{3}89432)$, but it doesn’t work for the second $(63.18^{\b{2}} = 3991.\b{7}124)$ where the first digit is 7 instead of 2.

Hint: The smallest such number is less than 10.

To prove that the problem is possible to solve quickly, I’ll provide the solution to a more complicated version. Suppose that the first two digits of $x^{j}$ must match $j$ instead of just the first one. This means that $x^{3}$ has “03” just to the right of the decimal point all way to up to $x^{99}$ having “99” just after the decimal point. Then the smallest solution is the following number:


The smallest solution’s digits actually never end so I had to cut it off somewhere. I choose the cut off point to be the first place where the remaining digits still worked. In other words, if you truncated the 4 off of the end, the number would be changed too much and would fail to meet all the conditions. Furthermore, I was forced to round up (because anything smaller doesn’t work).

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