<s>[INST] Find the largest number having this property. The $13$-digit number $1200549600848$ has the property that for any $1 \le n \le 13$, the number formed by the first $n$ digits of $1200549600848$ is divisible by $n$ (e.g. 1|2, 2|12, 3|120, 4|1200, 5|12005, ..., 13|1200549600848 using divisor notation).
Question 1: Find the largest computed number having this property.
Question 2: Is there a theoretical upper bound on the largest possible number with this property?
Edit: Added Question 2 as I believe it is more insightful as compared to brute force computer calculations.
 [/INST] Let $n_k = 10^{k - 1} a_1 + 10^{k - 2} a_2 + \cdots + a_k$ for positive integer $k$. Also,
$$\begin {eqnarray}
n_{k + 1} & = & 10^k a_1 + 10^{k - 1} a_2 + \cdots + 10 a_k + a_{k + 1} \nonumber \\ & = & 10 n_k + a_{k + 1}. \tag{1}
\end {eqnarray}$$
Since $k | n_k$, we have, by $(1)$, that
$$\begin {equation*}
\begin {cases}
k + 1 |10 n_k + a_{k + 1}, \\
k | n_k, \tag{2} \\ 
0 \leqslant a_{k + 1} \leqslant 9.
\end {cases}
\end {equation*}$$
Solving $(2)$ we have $$k + 1 | 10 r_k - a_{k + 1},$$ where $r_k = n_k/k$. Thus,
$$\begin {equation*}
\begin {cases}
a_k \equiv 10 r_{k - 1} \pmod {k}, \\
0 \leqslant a_k \leqslant 9. \tag{3} \\
\end {cases}
\end {equation*}$$
Let $S_k$ be the set of solutions of $(3)$. Then, the number of such numbers is $$\sum_{\substack {k \geqslant 1 \\ n_k \in S_k}} 1 = \sum_{k = 1}^{\infty} |S_k|.$$
Empirical data shows that for some $k$ the set $S_k$ is empty. We can even state a far stronger claim, that the set $S_k$ is empty for infinitely many $k$.
Lemma. There exists a natural number $N$ such that $S_k$ is empty for all $k \geqslant N$.
Proof. Let $N$ be the smallest integer such that $S_N$ is empty, that is, $n_N$ does not exist. Observe that if $S_N$ is empty, then all $S_k$ are also empty for $k > N$, since if $n_N$ does not exist, $n_k$ for $k > N$ cannot exist, either.
Necessary and sufficient condition for minimality of non-existence of $n_N$ is that $n_{N - 1}$ exists, but that $N$ never divides $10 r_{N - 1} - a_N$. We can guarantee that if and only if when we write $10 r_{N - 1} = N t + b$, where $t$ and $b$ are positive integers and $b \leqslant N - 1$, then $$10 \leqslant b \leqslant N - 1.$$ Indeed, then we would have $10 r_{N - 1} - a_N = N t + b - a_N$ and $0 \leqslant a \leqslant 9$, which gives $$N t + 1 \leqslant N t + b - a_N = 10 r_{N - 1} - a_N \leqslant N t + N - 1,$$ which is never divisible by $N$.
Suppose that $N$ is odd. Denote by $\langle x \rangle$ the smallest positive integer that a positive integer $x$ gives modulo $N$. Let $m$ be the smallest positive integer such that $10 m \equiv 1 \pmod {N}$, which exists, since $N$ is odd. Then, $10 m r_{N - 1} \equiv r_{N - 1} \pmod {N}$. On the other hand, we have $10 m r_{N - 1} \equiv m b \pmod {N}$. Hence, $$r_{N - 1} \equiv m b \pmod {N}.$$ Since $10 \leqslant b \leqslant N - 1$, we have $$1 = \langle 10 m \rangle \leqslant \langle m b \rangle = \langle r_{N - 1} \rangle \leqslant N - m.$$ Since $n_{N - 1} \equiv -r_{N - 1} \pmod {N}$, we further have $$m \leqslant \langle n_{N - 1} \rangle \leqslant N - 1.$$
to be continued...
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