This is my first time posting so do correct me if I am doing anything wrong.
Please help me with this math problem from the British Maths Olympiad (1994 British Maths Olympiad1 Q1 Number Theory).
Starting with any three digit number $n$ (such as $n = 625$) we obtain a new number $f(n)$ which is equal to the sum of the three digits of $n$, their three products in pairs, and the product of all three digits. Find all three digit numbers such that $\frac{n}{f(n)}=1$.
The only solution I found is $199$, can someone verify it please?