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
Topic
Inequality

Difficulty Level
9 out of 10

Suggested Book

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

Start with hints

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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