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 $\frac{n}{f(n)}=1$.
> 
> 
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The only solution I found is $199$, can someone verify it please?

Answer: 
Let $n=100a+10b+c,$ where $a> 0$ and $b,c\geq 0$. 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-9\leq 0$, but $b+c+bc\leq 99$. So the above equation holds iff $b=c=9$, which means $a$ can take any value.
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)