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}}