# 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.) .

# Connected Program at Cheenta

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.