2-Digit Numbers divisible by both Product and Sum of Digits

Theorem
The $2$-digit positive integers which are divisible by both the sum and product of their digits are:
 * $12, 24, 36$

Proof
We have:

It remains to be demonstrated that these are the only ones.

Let $z$ be a $2$-digit positive integer which is divisible by both the sum and product of its digits.

Neither of its digits may be $0$, because the product of the digits is a positive integer.

So, for $a \in \set {1, \dots, 9}$, $b \in \set {1, \dots, 9}$, $m \in \N$, $n \in \N$ we have:

Consider $(1) = (2)$:

In $(4)$ $a$, $b$, and $m - 1$ are all strictly positive; hence $10 - m$ must also be strictly positive.

Hence $1 < m < 10$.

Consider $(1) = (3)$:

Since $n \in \N$, we must have:
 * $\paren {m - 1} \divides 9 m$

Since $m$ and $m - 1$ are coprime, by Euclid's Lemma:
 * $\paren {m - 1} \divides 9$

This gives $m \in \set {2, 4}$.

Consider the case $m = 2$:

From $(4)$, $\dfrac b a = 8$.

Hence the only potential value for $b$ is $8$.

From $(5)$, $n = \dfrac {18} 8 \notin \N$.

Hence there are no potential candidates for $z$ for the case $m = 2$.

Consider the case $m = 4$:

From $(4)$, $\dfrac b a = 2$.

Hence the only potential values are $b \in \set {2, 4, 6, 8}$.

From $(5)$, $n = \dfrac {12} b$.

However, $8$ does not divide $12$.

This leaves $b \in \set {2, 4, 6}$.

By virtue of $\dfrac b a = 2$, we have $z \in \set {12, 24, 36}$.

We have now considered all the potential values of $m$.

Hence we conclude that the only $2$-digit positive integers which are divisible by both the sum and product of their digits are $12$, $24$, and $36$.