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## Understand the problem

Let $g : \mathbb{N} \to \mathbb{N}$ with $g(n)$ being the product of digits of $n$.        (a) Prove that $g(n)\le n$ for all $n \in \mathbb{N}$ .        (b) Find all $n \in \mathbb{N}$ , for which $n^2-12n+36=g(n)$.
##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 5 from 2017
Inequality

9 out of 10

##### Suggested Book

Challenge and Thrill of Pre-College Mathematics by V.Krishnamuthy , C.R.Pranesachar, ect.

Do you really need a hint? Try it first!

Let $n$ be a $k$ digit(s) number number , then $n$ can be written as                             $n=a_0+10a_1+10^2a_2+\cdots+10^{k-1}a_{k-1}$ Where ,$a_0,a_1,…,a_{k-1}$ are digits of $n$.

$a_0,a_1,…,a_{k-1} \in [1,9]$ as the range of the function $g$ is $\mathbb{N}$ $\Rightarrow a_0,a_1,…,a_{k-1}\neq 0$ . Now $g(n)=a_0a_1a_2\cdots a_{k-1}\le \underbrace{10\cdot10\cdot10\cdots 10}_{(k-1) times} \cdot a_{k-1}$   [Since $a_0,a_1,…,a_{k-1} \le 10$]  Equality holds when $k=1$ .

$\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}$ $\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}+\cdots+10^2a_2+10a_1+a_0$    [Since $a_0,a_1,…,a_{k-1} >0$]  $\Rightarrow g(n)\le n$        (Proved) .

$n^2-12n+36=g(n)$ $\Rightarrow n^2-12n+36 \le n$     [Since $g(n) \le n$ ] $\Rightarrow n^2-13n+36 \le 0$ $\Rightarrow (n-9)(n-4) \le 0$ $\Rightarrow 4\le n\le 9$ $\Rightarrow n={4,5,6,7,8,9}$ (Ans.)  .

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