Numbers Divisible by Sum and Product of Digits

Theorem
The sequence of positive integers which are divisible by both the sum and product of its digits begins:


 * $1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 111, 112, 132, 135, \ldots$

Proof
Let $S$ be the set of all positive integers which are divisible by both the sum and product of their digits.

Trivially, the sum and product of the digits of a one-digit number $n$ are themselves $n$.

Thus from Integer Divides Itself, the positive integers from $1$ to $9$ are in $S$.

The product of any integer with a $0$ in it is $0$.

From Zero Divides Zero $0$ is not a divisor of $n$ unless $n$ is itself $0$.

So $10, 20, 30, \ldots$ are not in $S$.

For all prime numbers $p$ with $2$ or more digits, the sum of its digits is greater than $1$ and less than $p$.

Thus $p$ is not a multiple of the sum of its digits.