Number equal to the product of its digits

Question

I'm trying to find a positive integer $n>10$ such that $n=p(n)$

Here $p(n)$ is defined as the product of the digits of $n$.

Example: $p(15)=1 \times 5=5$

I actually don't know where to start. What will be a good approach to this problem?

Answer 1

You will find none.
Proof Say that $K\ge10$ has $n$ digits, from left to right, $a_1,a_2,...a_n$. Then the number is $K=a_1*10^{n-1}+a_2*10^{n-2}+...+a_n*10^0$. Then $K\ge a_1*10^{n-1}$. Also note that $a_1\not=0$.

Also one wants $K=a_1*(a_2*a_3*...*a_n)$. However, since $0\le a_i\le9$ for all $i$, $(a_2*a_3*...*a_n)\le(9*9*9...*9)=9^{n-1}$. Therefore $K\le a_1*9^{n-1}$, which gives a contradiction.

Therefore there is no such $K\ge10$. $\square$