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 $$n=4,9$$.