How long can this string of digits be extended? Consider a number $a_1a_2a_3a_4 \dots a_n$ in some base $b$, such that for each $k, 1\leq k \leq n$, the subnumber $a_1a_2\dots a_k$ is a multiple of $k$.
For instance $1836$ is such a number in base $10$, because $1$ is a multiple of $1$, $18$ is a multiple of $2$, $183$ is a multiple of $3$, and $1836$ is a multiple of $4$.
Let $N(b)$ be the maximum possible value of $n$ for base $b$.

How large is $N(b)$?

We might expect $N(b)$  to be about $eb$. Indeed, there are about $b^n/n!$ such numbers of length $n$, which by Stirling's approximation goes below $1$ sometime around $n=eb$.
The only lower bound I have is that $N(b)\geq b$. I don't have any upper bound at all.
This question is the result of a conversation with John Conway.
 A: Following links at the OEIS entry mentioned above takes one in a step or two to this page where there are posts (from 2005) with  Maple code and results out to base $23$. The values $N(b)$ for $2 \le b \le 23$ are reported to be 
$2, 6, 7,  10, 11, 18, 17, 22, 25,  26, 28, 35, 39,  38, 39, 45, 48, 48,  52, 53, 56, 58$ 
Note that $N(7) \gt N(8)$ and $N(14) \gt N(15)$ and $N(19)=N(20).$
The ratios $\frac{N(b)}{b}$ are
$1.0, 2.0, 1.75, 2.0, 1.833, 2.571, 2.125, 2.444, 2.5, 2.364, 2.333, 2.692, $$2.786, 2.533, 2.438, 2.647, 2.667, 2.526, 2.60, 2.524, 2.546, 2.522$
Based on the data so far one might feel somewhat safe speculating that  $2\lt \frac{N(b)}{b} \lt 3$ provided $b \gt 6.$ As far as I can see, little nothing is known for sure (including that $N(b)$ is finite althoughthat seems highly likely.)
For each fixed value of $b$ there is a tree of possibilities (if we use a formal root node for level $0$.) A node at level $k-1$ has at most $\lceil \frac{b}{k}\rceil$ children. It might (or might not) be worth looking at the distribution of leaf levels.
