Q:

Decimal digits multiplied by powers of 2: leads to mod 8?

This is more a puzzle than a research question,
a puzzle to me. Perhaps it is straightforward for others.
Imagine Repeatedly interpreting a number
expressed with the usual base-$10$ digits
as "digits" multiplying powers of $2$ rather than powers of $10$.
For example, interpret
$n=27$ as $2\cdot 2^1 + 7\cdot 2^0 = 11$.
Call this mapping $f(n)$.
$f(n)=n$ when $n$ is a single (decimal) digit.
Let $g(n)$ be repeated application of $f(n)$,
$g(n) = f^k(n)$ until a single digit is reached.
For example, for
$g(17355)=3$:
$$
1\cdot 2^4 + 7\cdot 2^3 + 3\cdot 2^2 + 5\cdot 2^1 + 5\cdot 2^0 =
(1,7,3,5,5)\cdot(16,8,4,2,1) = 99
$$
$$
(9,9)\cdot(2,1) = 27
$$
$$
(2,7)\cdot(2,1) = 11
$$
$$
(1,1)\cdot(2,1) = 3
$$
$g(17356)=4$:
$$
(1,7,3,5,6)\cdot(16,8,4,2,1) = 100
$$
$$
(1,0,0)\cdot(4,2,1) = 4
$$
It is easy to see that $g(n)$ is well-defined, in that
it does eventually reach a single digit:
$f(n) < n$ for all $n \ge 10$
because $2^k < 10^k$ for $k \ge 1$.
It appears that
$g(n)$ is essentially $({n}\mod 8)$.
More precisely,
let $m= ({n}\mod 8)$.
Then I think that:
$$g(n) = m \;\; \mathrm{if} \;\; m > 1$$
$$g(n) = m+8 \;\; \mathrm{if} \;\; m \le 1$$
But I do not see a proof.

Q1. Is the above mod-$8$ formula for $g(n)$ correct?
  Is there a simple proof?
Q2. What is the generalization with
  base $10$ and base $2$ replaced
  with base $b_1$ and $b_2$, $b_1 > b_2$?

A:

Just a simple proof of the formula for $2$ and $10$. The remainder modulo $8$ is unchanged under the operation $f$, since $2^k=10^k$ modulo $8$. On the other hand, you will never reach $0$ or $1$ if you did not start with them. 

