Question: 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?

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

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We want to compute all possible integers $n$ such that $\frac{n}{f(n)} = 1$. 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 $\frac{n}{f(n)} = 1,$ 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

$$\boxed{\{199, 299, 399, 499, 599, 699, 799, 899, 999\}}$$
let n=100a+10b+c

f(n)= a+b+c+ab+bc+ca+abc
if n/f(n)=1 we get
n=f(n)

100a+10b+c=a+b+c+ab+bc+ca+abc  

99a+9b=ab+bc+ca+abc

99a+9b=a(b+c+ca)+bc  

Let a=1  

99+9b=b+c+2bc  

99+8b=c(1+2b)=(95+4)+8b= 95 + 4(1+2b)  

95=c(1+2b)-4(1+2b)=(c-4)(1+2b)  

since c, b are integers from [1,9]  

95=19\*5=(c-4)(1+2b)  

c-4 is single digit so it must be = 5  

c-4=5  

c=9  

1+2b=19  

b=9  

so abc becomes 199  

and similarly taking a values from 2 to 9...  

(this method is time consuming, I couldn't think of a shorter one)