Product of digits

Since $k > 0$ , we must have $P(k) > 0 \$ :

$\displaystyle \begin{aligned} k^2 - 10k - 22 & > 0 \\ k & > 11.86 \\ \implies k & \geq 12 \end{aligned}$

Furthermore, $P(n) \leq n \ \forall \ n \in \mathbb{N}$ . Let $d$ be the number of digits of $n$ , and $a > 0$ be the first digit of $n$ . The product of its digits is then bounded above by $a \times 9^{d - 1}$ , while the number $n$ itself is bounded below by $a \times 10^{d - 1}$ .

Thus, we have

$\displaystyle \begin{aligned} k^2 - 10k - 22 & \leq k \\ k & \leq 12.73 \\ \implies k & \leq 12 \end{aligned}$

We check that $k = 12$ is a solution, and see that it is indeed a (and thus the only) solution. Our answer is $\boxed{12}$ .