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

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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 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:

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


Consider $(1) = (2)$:

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

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

Hence $1 < m < 10$.


Consider $(1) = (3)$:

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

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 $36$.

$\blacksquare$


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