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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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