ISI Power

$n^2-12n+36=g(n) \\ g(n)=(n-6)^2$

We know product of digits of a number can not exceed the number itself

So, $n \geq g(n) \\ n-6 \geq g(n)-6 \\ (n-6)^2 \geq (g(n)-6)^2 \\ g(n) \geq (g(n))^2-12g(n)+36 \\ (g(n))^2-13g(n)+36 \leq 0 \\ (g(n)-4)(g(n)-9) \leq 0$

It implies $4 \leq g(n) \leq 9$

It can be easily checked and the only two solutions will be $4,9$

Let $n=\overline{a_k a_{k-1} ... a_{1}}$ be a $k$ -digit number, thus $n=10^{k-1}a_k+10^{k-2}a_{k-1}+...+10a_2+a_1$ Observe that $g(n)=a_1 a_2 .. a_{k-1} a_k \le 9^{k-1} a_k \le 10^{k-1} a_k \le 10^{k-1}a_k+10^{k-2}a_{k-1}+...+10a_2+a_1=n$ Therefore $g(n) \le n$ . Now for this problem we have $n^2-12n+36=g(n)\le n \Longrightarrow n^2-13n+36 \le 0 \Longrightarrow (n-4)(n-9) \le 0$ Hence $4 \le n \le 9$ . Equality occurs when $n=4, 9$ .