A Problem from the Australian Math Olympiad

Let $S(n)$ be the product of the digits of $n\in\mathbb{N}$. Find all natural numbers $n$ such that $S(n)=n^2-17n+56$.
*Hint: try to first prove that $n\geq S(n)$.

Lemma: For all $n\in\mathbb{N}$, we have that $n\geq S(n)$.
Let $n=\overline{x_kx_{k-1}\ldots x_1x_0}=x_k\cdot 10^k+x_{k-1}\cdot 10^{k-1}\ldots+x_1\cdot 10+x_0$. Then we have that $$S(n)=x_kx_{k-1}\cdots x_1x_0=\prod_{i=0}^{k}x_i.$$ Since each $x_i\in\{0,1,\ldots,9\}\Rightarrow x_i\leq 9$, we can establish the desired result as follows: $$\begin{equation} \begin{split} S(n)&=x_kx_{k-1}\cdots x_1x_0\\ &\leq x_k\cdot 9^k\\ &\leq x_k \cdot 10^k\\ &\leq x_k \cdot 10^k+x_{k-1}\cdot 10^{k-1}\ldots+x_1\cdot 10+x_0=n.\\ \end{split} \end{equation}$$ This proves the result that $n\geq S(n)$ for all natural numbers $n$. $\blacksquare$

Using the result of the above lemma, we can now establish an ineqaulity in terms of one variable: $$S(n)=n^2-17n+56\leq n\Leftrightarrow n^2-18n+56=(n-4)(n-14)\leq 0\Leftrightarrow 4\leq n \leq 14.$$ Therefore, it suffices to check if the condition is satsifed for the natural numbers from $4$ to $14$. Stragithforward computations will prduce the follwoing table: $$\begin{array}{|c|c|c|} \hline n & n^2-17n+56 & S(n)\\ \hline 4 & 4 & 4\\ \hline 5 & -4 & 5\\ \hline 6 & -10 & 6\\ \hline 7 & -14 & 7\\ \hline 8 & -16 & 8\\ \hline 9 & -16 & 9\\ \hline 10 & -14 & 0\\ \hline 11 & -10 & 1\\ \hline 12 & -4 & 2\\ \hline 13 & 4 & 3\\ \hline 14 & 14 & 4\\ \hline \end{array}$$ A simple way of getting the above values would have been to note that the $x$-value of the vertex of the parabola defined by $y=x^2-17x+56$ is given by $-\frac{b}{2a}=\frac{17}{2}=8.5$. Therefore, the value of $n^2-17n+56$ would be symmetric about the point $x=8.5$. From the table above, we can find that the only natural number that satsifes $S(n)=n^2-17n+56$ is $\boxed{n=4}$.