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 any 2-digit positive integer which is divisible by both the sum and product of its digits.

Neither of its digits may be zero as the product of the digits is a positive integer. So, for $ a \in \{ 1, ..., 9 \}$, $ b \in \{ 1, ..., 9 \} $, $ m \in \N $, $ n \in \N $ we have:

Consider $(1) = (2)$

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

Consider $(1) = (3)$

In $(5)$ for $ n \in \N $ we must have all three divisibility requirements met:
 * $ b \paren { m - 1 } \mid 9m $
 * $ b \mid 9m $
 * $ \paren { m - 1 } \mid 9m $

The last of the above three requirements only depends upon the value of $m$ and is therefore simpler to use to determine potential values of $m$ that may allow $ n \in \N $:
 * The only value of $m$ where $ \paren { m - 1 } \mid m$ is $m = 2$; hence $m = 2$ or $ \paren { m - 1 } \mid 9m$
 * If $m$ is odd then $ \paren { m - 1 } $ is even. Both $9$ and $m$ are odd, so $9m$ is also odd. Clearly, an even number cannot divide $9m$; therefore $m$ cannot be odd.
 * If $m$ is even then $ \paren { m - 1 } $ is odd and must divide $9m$. We know that the only value of $m$ where $ \paren { m - 1 } \mid m$ is $m = 2$, so any other cases where $ \paren { m - 1 } $ is odd must divide $9$. Hence $ \paren { m - 1 } \in \{ 1, 3 \} $
 * The potential values are therefore $ m \in \{ 2, 4 \} $

Consider the case $m = 2$:


 * From $(4) \frac b a = 8 $; hence the only potential value for $b$ is $8$.


 * From $(5) n = \frac { 18 } 8 \notin \N $; hence there are no potential candidates for $z$ for the case $m = 2$

Consider the case $m = 4$:


 * From $(4) \frac b a = 2 $; hence the only potential values are $ b \in \{ 2, 4, 6, 8 \} $


 * From $(5) n = \frac { 12 } b$; however, 8 does not divide 12 and therefore this leaves $ b \in \{ 2, 4, 6 \} $


 * By virtue of $\frac b a = 2 $ we therefore have for the case $m = 4$ that $ z \in \{ 12, 24, 36 \} $

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