Question 101
Let $n \geqslant 1$ be an integer. Morgane writes on the board, in base 10, the numbers $2023, 2023 \times 2, \ldots, 2023 \times n$. For every digit $c$ between 1 and 9, she then denotes by $\mathrm{d}_{c}(n)$ the number of occurrences of the digit $c$ on the board. For example, if $n=3$, she writes the numbers $2023$, $4046$ and $6069$, so $\mathrm{d}_{1}(3)=\mathrm{d}_{5}(3)=\mathrm{d}_{7}(3)=\mathrm{d}_{8}(3)=0$, $\mathrm{d}_{3}(3)=\mathrm{d}_{9}(3)=1$, $\mathrm{d}_{2}(3)=\mathrm{d}_{4}(3)=2$ and $\mathrm{d}_{6}(3)=3$; these nine numbers therefore take exactly four values.
Prove that there exist infinitely many integers $n \geqslant 1$ for which the nine numbers $\mathrm{d}_{1}(n), \mathrm{d}_{2}(n), \ldots, \mathrm{d}_{9}(n)$ take exactly two values.
Prove that there exist infinitely many integers $n \geqslant 1$ for which the nine numbers $\mathrm{d}_{1}(n), \mathrm{d}_{2}(n), \ldots, \mathrm{d}_{9}(n)$ take exactly two values.