MMO2025 Round 4

Answer: 4050 Put $n=2025$ and $d=10^n+3$. Let $N=\overline{a_m\dots a_2{a}_1}$ be a multiple of $d$ such that the number of different digits in $N$ is at most two. Then we can assume that $a_1\ne 0$ and clearly $m\ge n+1$ since $N\ge d$. Suppose that $2n>m$, and consider $A=\overline{a_n\dots a_1}$ and $B=\overline{a_m\dots a_{n+1}}$. Since $d>3\times 10^{n-1}>3B$ and $d>A$ we have $|A-3B|<d$. Moreover, $A-3B=N-dB$ is a multiple of $d$ and so $A=3B$, or equivalently, $$\overline{a_n\dots a_1}=3\times \overline{a_m\dots a_{n+1}}.(0.2)$$ If $a_n=0$ then $a_{n+1}=a_1$ since $a_{n+1}\ne 0$ and $a_1\ne 0$. Therefore, $a_1=5$ and $3a_{n+2}+1\equiv a_2(\mathrm{mod}10)$ by (0.2). But this is impossible since $a_2,a_{n+2}\in \{0,5\}$. Thus $a_n\ne 0$ and so it follows from (0.2) that $m=2n-1$. Moreover, we have $\overline{a_n{a}_{n-1}}=3a_m+c$ for some digit $c$. Observe that $1\le a_n\le 2$ and $0\le c\le 2$ since $3B<3\times 10^{n-1}$ and $3\times \overline{a_{m-1}\dots a_{n+1}}<3\times 10^{n-2}$. Assume that $a_n=2$. If $a_1=2$ then $a_{n+1}=4$ by (0.2). Hence $c=\overline{a_n{a}_{n-1}}-3a_m\ge 22-3\times 4=10$, a contradiction. If $a_{n+1}=2$ then $a_1=6$ and so $c\ge 22-3\times 6=4$, a contradiction. Assume that $a_n=1$. If $a_{n+1}=1$ then $a_1=3$, hence (0.2) is impossible. If $a_1=1$ then $a_{n+1}=7$, hence $c\ge 11-3\times 1=8$, a contradiction. Thus $m\ge 2n$ and so it suffices to show that $2n$-digit number $\overline{1\dots 113\dots 33}$ is divisible by $d$: $$\begin{array}{rl}\overline{1\dots 113\dots 33}& =\overline{1\dots 11}\times 10^n+\overline{3\dots 33}\\ & =\overline{1\dots 11}\times (d-3)+3\times \overline{1\dots 11}\equiv 0(\mathrm{mod}d).\end{array}$$