# 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:

 $\displaystyle 12$ $=$ $\displaystyle 4 \times \paren {1 + 2}$ $\displaystyle$ $=$ $\displaystyle 6 \times \paren {1 \times 2}$
 $\displaystyle 24$ $=$ $\displaystyle 4 \times \paren {2 + 4}$ $\displaystyle$ $=$ $\displaystyle 3 \times \paren {2 \times 4}$
 $\displaystyle 36$ $=$ $\displaystyle 4 \times \paren {3 + 6}$ $\displaystyle$ $=$ $\displaystyle 2 \times \paren {3 \times 6}$

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:

 $(1):\quad$ $\displaystyle z$ $=$ $\displaystyle 10a + b$ 2-digit positive integer $(2):\quad$ $\displaystyle z$ $=$ $\displaystyle \paren { a + b } m$ divisible by the sum of its digits $(3):\quad$ $\displaystyle z$ $=$ $\displaystyle abn$ divisible by the product of its digits

Consider $(1) = (2)$

 $\displaystyle 10a + b$ $=$ $\displaystyle \paren { a + b } m$ Note: if $m = 1$ then $a = 0$; hence $m > 1$ $\displaystyle 10a - am$ $=$ $\displaystyle bm - b$ rearranging $\displaystyle a \paren { 10 - m }$ $=$ $\displaystyle b \paren { m - 1 }$ factorizing $(4):\quad$ $\displaystyle \frac b a$ $=$ $\displaystyle \frac { 10 - m } { m - 1 }$ rearranging; $a > 0$ and $m > 1$ so both denominators are never zero

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

Consider $(1) = (3)$

 $\displaystyle 10a + b$ $=$ $\displaystyle abn$ $\displaystyle n$ $=$ $\displaystyle \frac { 10a + b } { ab }$ rearranging $\displaystyle$ $=$ $\displaystyle \frac { 10 + \frac b a } b$ dividing numerator and denominator by $a$, which is > 0 $\displaystyle$ $=$ $\displaystyle \frac { 10 + \frac { 10 - m } { m - 1 } } b$ from $(4)$ $\displaystyle$ $=$ $\displaystyle \frac { 10 \paren { m - 1 } + 10 - m } { b \paren { m - 1 } }$ multiplying numerator and denominator by $m - 1$, which is > 0 $(5):\quad$ $\displaystyle n$ $=$ $\displaystyle \frac { 9 m } { b \paren { m - 1 } }$ simplifying

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.

$\blacksquare$