The fear of Lord Voldemort (The Dark Lord)

Note that if $N = \overline{a_1a_2\cdots a_n}$ then the product of the digits of $N$ is $P(N) \; = \; a_1\times a_2 \times \cdots\times a_n \; \le \; a_1 \times 10^{n-1} \; \le \; \overline{a_1a_2\cdots a_n} \; = \; N$ Thus, if $N^2 - 10N - 22 = P(N)$ , then $\begin{aligned} N^2 - 10N - 22 & = \; P(N) \; \le \; N \\ N^2 - 11N - 22 & \le \; 0 \\ \tfrac12\big(11 - \sqrt{209}\big) & \le \; N \; \le \; \tfrac12\big(11 + \sqrt{209}\big) \end{aligned}$ and hence $1 \le N \le 12$ . Checking these values, the only natural number $N$ such that $N^2 - 10N - 22 = P(N)$ is $N = \boxed{12}$ .