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:
\(\ds 12\) | \(=\) | \(\ds 4 \times \paren {1 + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 6 \times \paren {1 \times 2}\) |
\(\ds 24\) | \(=\) | \(\ds 4 \times \paren {2 + 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times \paren {2 \times 4}\) |
\(\ds 36\) | \(=\) | \(\ds 4 \times \paren {3 + 6}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times \paren {3 \times 6}\) |
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:
\(\text {(1)}: \quad\) | \(\ds z\) | \(=\) | \(\ds 10 a + b\) | $2$-digit positive integer. | ||||||||||
\(\text {(2)}: \quad\) | \(\ds z\) | \(=\) | \(\ds \paren {a + b} m\) | divisible by the sum of its digits | ||||||||||
\(\text {(3)}: \quad\) | \(\ds z\) | \(=\) | \(\ds a b n\) | divisible by the product of its digits |
Consider $(1) = (2)$:
\(\ds 10 a + b\) | \(=\) | \(\ds \paren {a + b} m\) | Note that if $m = 1$ then $a = 0$; hence $m > 1$ | |||||||||||
\(\ds 10 a - a m\) | \(=\) | \(\ds b m - b\) | rearranging | |||||||||||
\(\ds a \paren {10 - m}\) | \(=\) | \(\ds b \paren {m - 1}\) | factorizing | |||||||||||
\(\text {(4)}: \quad\) | \(\ds \frac b a\) | \(=\) | \(\ds \frac {10 - m} {m - 1}\) | 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 $(1) = (3)$:
\(\ds 10 a + b\) | \(=\) | \(\ds a b n\) | ||||||||||||
\(\ds n\) | \(=\) | \(\ds \frac {10 a + b} {a b}\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {10 + \frac b a} b\) | dividing numerator and denominator by $a > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {10 + \frac {10 - m} {m - 1} } b\) | from $(4)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {10 \paren {m - 1} + 10 - m} {b \paren {m - 1} }\) | multiplying numerator and denominator by $m - 1 > 0$ | |||||||||||
\(\text {(5)}: \quad\) | \(\ds n\) | \(=\) | \(\ds \frac {9 m} {b \paren {m - 1} }\) | 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 $36$.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $36$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $36$