# Number theoretic function (I.S.I. B.Stat, B.Math 2017 Subjective 5)

Problem

Let $g: \mathbb{N} \to \mathbb{N}$ with g(n) being the product of the digits of n.

1. Prove that $g(n) \le n$ for all $n \in \mathbb{N}$
2. Find all $n \in \mathbb {N}$ for which $n^2 -12n + 36 = g(n)$

DIscussion

(1)Let,$$n=\sum_{i=0}^{k}10^{i}a_{i}.$$

Now,to prove that $g(n)\leq n$,we have to show that ,$$\sum_{i=0}^{k}10^{i}a_{i}\ge \prod_{j=0}^{k}a_{i}.$$

Let us define the statement ,$$P(k):\sum_{i=0}^{k}10^{i}a_{i}\ge \prod_{j=0}^{k}a_{i}.$$

Now,$P(0)$,$P(1)$ is true for $k=0$,$k=1$,respectively because

$a_{1}\in\{0,1,2,….,9\}$,which implies,that $10a_{1}\ge a_{1}$

Now by induction hypothesis,

Let us assume that $P(k)$ is true for some $k=m$,such that,

$$\sum_{i=0}^{m}10^{i}a_{i}\ge \prod_{j=0}^{m}a_{i}.$$

Now,we have $a_{m+1}>a_{m+1}-1$ and $10>a_{0},10>a_{1},\dots,10>a_{m}$

Multiplying all of these we get $$10^{m+1}a_{m+1}>(a_{m+1}-1)a_{0}a_{1}\dots a_{m}=(a_{m+1}-1)\prod_{j=0}^{m}a_{j}.$$

$\Rightarrow 10^{m+1}a_{m+1}+\sum_{i=0}^{m}10^{i}a_{i} \ge(a_{m+1}-1)a_{0}a_{1}\dots a_{m}+ \prod_{j=0}^{m}a_{i}\\ =(a_{m+1}-1)\prod_{j=0}^{m}a_{j}+\prod_{j=0}^{m}a_{j} \\ \Rightarrow \sum_{i=0}^{m+1}10^{m+1}a_{m+1}\ge\prod_{j=0}^{m+1}a_{j}$

Hence,$P(m+1)$ is true.

So,$P(k)$ is true for all $k\in\mathbb{N}\cup\{0\}$.

Thus,$g(n)\leq n$,for all $n\in\mathbb{N}$.

(2)Using $g(n)\leq n$,we have $$n^2-12n+36\leq n$$.

$$=>n^2-13n+36=(n-4)(n-9)\leq0=>n\in[4,9].$$

Now,since $n\in\mathbb{N}$,$n=4,5,6,7,8,9$.

Also,$n^2-12n+36=(n-6)^2=g(n)$,which implies $g(n)$ is a perfect square.

Since,$n<10$,$g(n)=n$,which implies $n$ is a perfect square.

But $n\in[4,9]$ and hence only solutions are $n=4,9$.

## 2 Replies to “Number theoretic function (I.S.I. B.Stat, B.Math 2017 Subjective 5)”

1. Rishabh says:

Isn’t the proof of the first part overkill? Why not just use $latex a_n10^n+ \cdots +a_0 \geq a_n10^n \geq a_n \cdots a_0$, since $latex 0<a_0, \cdots a_{n-1}<10$? The proof is both shorter and simple.

1. Anish Ray says:

Correct…the concept used is same though.