MMO2025 Round 4

Answer: $2n+1$ Let $N=\overline{a_m\dots a_2{a}_1}$ be a multiple of $d$ such that $a_i\in \{1,2\}$ for every $1\le i\le m$, where $d$ denotes $10^n-7$. Since $N\ge 2d$ it is clear that $m\ge n+1$. Suppose that $2n\ge m$. By setting $S=7\times \overline{a_m\dots a_{n+1}}+\overline{a_n\dots a_1}$ we get $$S\le 7\times \underset{n}{\underbrace{22\dots 2}}+\underset{n}{\underbrace{22\dots 2}}<2d.$$ It follows from $d\mid N$ that $d\mid S$ and so we must have $S=d$ $\left(\ast )\right)$. But this equality is impossible. Indeed, by $\left(\ast )\right)$, we have $S\equiv 7a_{n+1}+a_1(\mathrm{mod}10)$ which implies $S\not\equiv 3(\mathrm{mod}10)$ since $a_1,a_{n+1}\in \{1,2\}$. This contradicts to $\left(\ast )\right)$. Thus $m\ge 2n+1$ and so, for our purpose, it suffices to show that $2\underset{n-1}{\underbrace{1\dots 11}}\underset{n-2}{\underbrace{2\dots 22}}111$ is divisible by $d$:

$$\begin{array}{rl}2\underset{n-1}{\underbrace{1\dots 11}}\underset{n-2}{\underbrace{2\dots 22}}111& =2\times 10^{2n}+\underset{n-1}{\underbrace{1\dots 11}}2\times 10^n+\underset{n-3}{\underbrace{2\dots 2}}111\\ & \equiv 2\times 7^2+7\times \underset{n-1}{\underbrace{1\dots 11}}2+\underset{n-3}{\underbrace{2\dots 2}}111\\ & \equiv \underset{n-2}{\underbrace{7\dots 7}}84+\underset{n-2}{\underbrace{2\dots 2}}09=d\equiv 0(\mathrm{mod}d).\end{array}$$