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?
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.