# Counting such 3-digit numbers

Count the number of 3-digit natural numbers N with the property that the sum of the digits of N is divisible by the product of the digits of N.

Let abc be the number then (abc)k = a + b + c

k= 1/bc+ 1/ac+ 1/ab

k is required to be a natural number for this to hold true

How do I proceed?

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 10 '18 at 14:32
• 1. None of the digits can be $0$. 2. Let $d$ be the largest of the digits. Then $a + b + c \leqslant 3d$, so the product of the other two digits is $\leqslant 3$, hence at least one of the digits is $1$. And $d$ divides the sum of the other two digits. If that sum is $2d$, i.e. all three digits are equal, we get $d^3 \mid 3d$, whence $d^2 \mid 3$, so $d = 1$. Otherwise, the sum of the two smaller digits is $d$, so the smaller digits are $1$ and $d-1$. Then we get $d(d-1) \mid 2d$, i.e. $(d-1)\mid 2$. – Daniel Fischer Aug 12 '18 at 15:19

So, we know that $\exists k\in\mathbb{N}$ such that $$(a+b+c) = k\cdot abc$$ and $$a\neq0$$Note that when $b,c = 0$, this holds since $abc=0$. Other than that, there isn't much more you can do than smart counting for this problem.
• Note: For the sum to be divisible by the product, you want $a+b+c=kabc$ for some integer $k$. – paw88789 Aug 10 '18 at 16:16