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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 $fracnf(n)=1$.

The only solution I found is $199$, can someone verify it please?

Let $n=100a+10b+c,$ where $a> 0$ and $b,cgeq 0$. We are trying to solve $$100a+10b+c=a+b+c+ab+ac+bc+abc \ implies 99a+9b=abc+ab+ac+bc \ implies a(99-b-c-bc)=b(c-9) \$$$c-9leq 0$, but $b+c+bcleq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.

Here's part $(b)$ because I'm assuming you don't need help with part $(a)$:

We want to compute all possible integers $n$ such that $fracnf(n) = 1$. Since we know that, by assumption, $n$ is a three-digit number, we can write

$$n = 100a + 10b + c,$$

where $a, b, c$ are integers. If this is the case, in terms of our newly defined variables $a$, $b$, and $c$, we can express $f(n)$ as follows:

$$f(n) = abc + ab + bc + ac + a + b + c.$$

Now, in order to have $fracnf(n) = 1,$ we must have $n = f(n)$. This happens when

$$99a + 9b = abc + ab + bc + ac$$

$$Longleftrightarrow (9-c)b = a(bc + b + c - 99) $$

Also, we must have $b, c leq 9,$ which implies $bc + b + c - 99 leq 0$. However, since $a neq 0$ (if $a = 0$, we would be able to form a two or one-digit number instead of a three-digit one!), we conclude $b = c = 9$. Therefore, our solution set is given by

$$boxed199, 299, 399, 499, 599, 699, 799, 899, 999$$