you can think of it like just converting to base 9 (because there are 9 digits) but with different namings/symbols.

In base 10, you can decompose numbers by their decimal cases, for example $832 = 8\cdot 10^2 + 3\cdot 10^1 + 2\cdot 10^0$. The same goes for an arbitrary base $B > 1$.

Let $n$ be a number in base $B$ with digits $d_k d_{k-1} \ldots d_1 d_0$. Following the same logic as of base $10$, $n$ can be written as $d_k\cdot B^k + d_{k-1} \cdot B^{k-1} + \ldots + d_0 \cdot B^0$. Now, notice that, all digits have "values" divisible by $B$, except for digit $0$. So, we can write $n$ in the follow way $n = B(d_k\cdot B^{k-1} + d_{k-1} \cdot B^{k-2} + \ldots + d_1 \cdot B^0) + d_0$. By Euclid's division lemma, the quotient of the division of $n$ by $B$ is $d_k\cdot B^{k-1} + d_{k-1} \cdot B^{k-2} + \ldots + d_1 \cdot B^0$, and the remainder is $d_0$. Observe that the quotient of that division is the base $B$ number $d_k d_{k-1} \ldots d_1$, which is our original number $n$, without it's last digit. Therefore, we can repeat this procedure until we find all the digits of $n$.

In the problem you mentioned, you're basically counting in base $9$, except we are using the digits $0, 1, 2, 3, 5, 6, 7, 8, 9$ instead of $0$ to $8$. So, all you need to do is find the base $9$ number, and map the digits $0, 1, \ldots 8$ to $0, 1, 2, 3, 5, 6, 7, 8, 9$.