Your problem sets out to solve the following system of modular equations:
x1 + 2*x2 + 2*x3 + 2*x4 + 2*x5 + x6 + x7 = A (mod 100)
x4 + x5 = B (mod 10)
x2 + x3 = C (mod 10)
The first equivalence comes from steps 1-3 above, the second from step 4 and the third from step 5. The basic math that you need to solve a system of equivalences is the Chinese Remainder Theorem.
This system will not have a unique solution. To see how redundant it is, count the possibilities for the sequence (x1,x2,x3,x4,x5,x6,x7) and (C,B,A) separately. For the former, x1,x6,x7 are interchangeable as are x4,x5 and x2,x3, so we count them separately. There are (45+3-1 choose 3) ways to choose x1,x6,x7 each between 01 and 45 (remember to choose with repetition), and (45+2-1 choose 2) ways to choose each of the other pairs. This gives a total of
(47 choose 3)*(46 choose 2)^2 = 16 782 525
ways of choosing the xs. For the 4-digit sequence, there are 10 000 possibilities. The number above is far larger, so certainly there is a lot of repetition.
If you want to pin down just one x sequence that yields a given CBA sequence, then you can simplify the problem by choosing arbitrary values for x6,x7,x5,x3 and then using the CRT to solve for x1,x4,x2 (which may or may not exist).
For example, suppose you take
x3 = 19, x5 = 22, x6 = 31, x7 = 13
Then you solve for the remaining xs by solving this system
x4 + 22 = 3 (mod 10)
x2 + 19 = 4 (mod 10)
x1 + 2*x2 + 38 + 2*x4 + 44 + 31 + 13 = 82 (mod 100)
The first two equations are easy, though the solutions are not unique
x4 = 1 (mod 10) => x4 = 1, 11, 21, 31, or 41
x2 = 5 (mod 10) => x2 = 5, 15, 25, 35, or 45
Adding in the final equation to solve for x1 and pin down x2,x4 gives
x1 + 2*(x2 + x4) = 56 (mod 100)
which has many solutions, among them x1 = 4, x2 = 5, x4 = 21, but also x1 = 44, x2 = 5, x4 = 1 as well as many others.
The number $0.010010001000010000010000001\ldots$ is irrational and has no instances of any digit other than $0$ and $1$. More generally, pick any two digits $d$ and $e$, and form the number
$$0.\mathrm{dedeedeeedeeeedeeeeed}\dots\;;$$
it never becomes periodic, so it’s irrational. Of course one can insert any finite string of digits between the decimal point and the expansion of an irrational number and still have an irrational number, e.g., $0.345345010010001000010000010000001\ldots$.
Best Answer
So, you are looking for a way to flip the number about the decimal point in whatever base. Here is the function which will do it
$$f(x)=\sum_{n=0}^{\infty} 10^{-n-1} \left(\left\lfloor\frac{x}{10^n}\right\rfloor \mod 10\right)+\sum_{n=1}^{\infty} 10^{n-1} \left(\left\lfloor 10^nx\right\rfloor \mod 10\right)$$
which is just a compact form of an algorithm. This is basically starting from the decimal point, first going to the left the digits are flipped. Then the second sum starts from the decimal point and going to the right flips the digits.
For whatever base, just replace all of the 10's with whatever base you want. This will work fine when the number (in whatever base) has a terminated expansion on both sides of the decimal point meaning both sums are finite. For irrational numbers for example, the second sum won't converge and the function isn't defined at all so don't use this to flip the digits of $\sqrt{2}$ because it can't be done. Another way to say that is the algorithm won't terminate in finite time (and we don't have infinite memory) because we don't know the "last" digit of $\sqrt{2}$ in base 10 for example. And even for some rational numbers this function doesn't make sense. For example you can't flip the decimal expansion of 1/3 in base 10 because it doesn't terminate in base 10.