Subject (3). For any natural number $n$, we denote by $P_{n}$ the product of its digits.

a) Show that for all natural numbers $n$ with 2013 digits, we have $P_{n}<10^{2013}$. . ...................................................................................

b) Find all natural numbers $n$ such that $P_{n}=\frac{25}{8} n-211$. . . . 5 points
Solution.

a) We have $P_{n} \leq 9^{2013}<10^{2013}$. Duh ... (probably for b)).

b) Let $n=\overline{a d_{1} d_{2} \ldots d_{k}}, k \in \mathbb{N}$. Then $n \geq 10^{k} a$, and $P_{n} \leq 9^{k} a \leq 10^{k} a$, so $10^{k} a \geq P_{n}=\frac{25}{8} n-211 \geq \frac{25}{8} 10^{k} a-211$, thus $10^{k} \leq \frac{8 \cdot 211}{17 a}<100$, which means $0 \leq k \leq 1$. On the other hand, we need $8 \mid n$, so $n=8 m$; now having $0 \leq P_{8 m}=25 m-211$ forces $m \geq\lceil 211 / 25\rceil=9$.

The only possibilities are thus $m=9$, for which $14=P_{72}=\frac{25}{8} \cdot 72-211$

and $m=11$, for which $64=P_{88}=\frac{25}{8} \cdot 88-211$.