Numbers Divisible by Sum and Product of Digits
Theorem
The sequence of positive integers which are divisible by both the sum and product of its digits begins:
- $1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 24, 36, 111, 112, 132, 135, \ldots$
This sequence is A038186 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
Let $S$ be the set of all positive integers which are divisible by both the sum and product of their digits.
Trivially, the sum and product of the digits of a one-digit number $n$ are themselves $n$.
Thus from Integer Divides Itself, the positive integers from $1$ to $9$ are in $S$.
The product of any integer with a $0$ in it is $0$.
From Zero Divides Zero $0$ is not a divisor of $n$ unless $n$ is itself $0$.
So $10, 20, 30, \ldots$ are not in $S$.
For all prime numbers $p$ with $2$ or more digits, the sum of its digits is greater than $1$ and less than $p$.
Thus $p$ is not a multiple of the sum of its digits.
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Sources
- Feb. 1990: Richard K. Guy: The Second Strong Law of Small Numbers (Mathematics Magazine Vol. 63, no. 1: pp. 3 – 20) www.jstor.org/stable/2691503
- This citation is given in Curious and Interesting Numbers, 2nd ed., but it is doubtful.
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $24$