MMO2025 Round 4

Assume to the contrary that there exists an integer $A=\overline{a_m\dots a_1}$ such that $\overline{a_m\dots a_1}$ has at most 2 different digits and $d\mid A$, where $d$ denotes $10^n-3$. Consider $S=\overline{a_n\dots a_1}+3\times \overline{a_m\dots a_{n+1}}$ which exists since $m>n$. It follows from $d\mid A$ that $d\mid S$. Moreover, the assumption that $2n-2\ge m$ yields $S\le 10^n-1+3\times (10^{n-2}-1)<2d$ and so we must have $$\overline{a_n\dots a_1}+3\times \overline{a_m\dots a_{n+1}}=10^n-3.(0.1)$$ Hence $a_1+3a_{n+1}\equiv 7(\mathrm{mod}10)$ and consequently, $a_1\ne a_{n+1}$. Observe that $$\overline{a_n\dots a_1}=d-3\times \overline{a_m\dots a_{n+1}}>d-3\times (10^{n-2}-1)>9\times 10^{n-1},$$ which implies $a_n=9$. Thus $9\in \{a_1,a_{n+1}\}$ by the assumption on $A$. If $a_{n+1}=9$ then $a_1=0$ and so $A$ is divisible by 9, contradicting to (0.1). If $a_1=9$ then $a_{n+1}=6$, giving a similar contradiction.