1. Determine the non-zero digits $a, b$ and $c$ for which $\frac{1}{a+b+c}=\overline{0.abc}$. Find all solutions!
Solution. By multiplying the given equation by $1000(a+b+c)$, we get $1000=\overline{a b c}(a+b+c)$. Since $a, b$, and $c$ are digits, for them it holds that $a+b+c \leq 27$. On the other hand, we have that 1000 as a product of two numbers (one of which is a three-digit number) can be represented as 500$\cdot$2, 250$\cdot$4, 200$\cdot$5, 125$\cdot$8, 100$\cdot$10. We check for which of the three-digit numbers the sum of the digits gives the second factor. This is only true for $125 \cdot 8$, so $a=1, b=2$ and $c=5$.