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

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

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 12
      | r = 4 \times \paren {1 + 2}
      | c = 
}}
{{eqn | r = 6 \times \paren {1 \times 2}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 24
      | r = 4 \times \paren {2 + 4}
      | c = 
}}
{{eqn | r = 3 \times \paren {2 \times 4}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 36
      | r = 4 \times \paren {3 + 6}
      | c = 
}}
{{eqn | r = 2 \times \paren {3 \times 6}
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | n = 1
      | l = z
      | r = 10 a + b
      | c = $2$-digit positive integer.
}}
{{eqn | n = 2
      | l = z
      | r = \paren {a + b} m
      | c = divisible by the sum of its digits
}}
{{eqn | n = 3
      | l = z
      | r = a b n
      | c = divisible by the product of its digits
}}
{{end-eqn}}
Consider $(1) = (2)$:
{{begin-eqn}}
{{eqn | l = 10 a + b
      | r = \paren {a + b} m
      | c = Note that if $m = 1$ then $a = 0$; hence $m > 1$
}}
{{eqn | l = 10 a - a m
      | r = b m - b
      | c = rearranging
}}
{{eqn | l = a \paren {10 - m}
      | r = b \paren {m - 1}
      | c = factorizing
}}
{{eqn | n = 4
      | l = \frac b a
      | r = \frac {10 - m} {m - 1}
      | c = rearranging; $a > 0$ and $m > 1$ so both denominators are never zero
}}
{{end-eqn}}
In $(4)$ $a$, $b$, and $m - 1$ are all strictly positive; hence $10 - m$ must also be strictly positive.
Hence $1 < m < 10$.
Consider $(1) = (3)$:
{{begin-eqn}}
{{eqn | l = 10 a + b
      | r = a b n
      | c = 
}}
{{eqn | l = n
      | r = \frac {10 a + b} {a b}
      | c = rearranging
}}
{{eqn | r = \frac {10 + \frac b a} b
      | c = dividing numerator and denominator by $a > 0$
}}
{{eqn | r = \frac {10 + \frac {10 - m} {m - 1} } b
      | c = from $(4)$
}}
{{eqn | r = \frac {10 \paren {m - 1} + 10 - m} {b \paren {m - 1} }
      | c = multiplying numerator and denominator by $m - 1 > 0$
}}
{{eqn | n = 5
      | l = n
      | r = \frac {9 m} {b \paren {m - 1} }
      | c = simplifying
}}
{{end-eqn}}
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 $36$.
{{qed}}