Given a three digit number $n$, let $f(n)$ be the sum of digits of $n$, their products in pairs, and the product of all digits. When does $n=f(n)$?

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?

• in the definition of $f$, are you concatenating the three results? Can you give an example of a pair $(n,f(n))$? – mathworker21 Nov 19 '18 at 5:44
• So $f(199) = 19, 9918, 81$? – steven gregory Nov 19 '18 at 5:45
• @mathworker21`@stevengregory For example if $n=625$, $f(n)=6+2+5+6*2+6*5+2*5+6*2*5$ – 3684 Nov 19 '18 at 5:51
• to do this problem, write $n = 100a+10b+c$ and just write everything out and solve the equation you get – mathworker21 Nov 19 '18 at 5:56
• @mathworker21, I have tried that but I didn't get too far, can you try it if you have time? I also thought about factorising a+b+c+ab+ac+bc+abc as (a+1)(b+1)(c+1)-1 but still wasn't able to get far. For number theory Olympiad problems are there systematic methods or does it require a different method every time. – 3684 Nov 19 '18 at 6:00

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.

• May I ask how you got to the solution so quick, do you just see the solution? – 3684 Nov 19 '18 at 6:13
• @3684 this problem is fairly 'routine'. Plus, I might've seen this before/its associated solution, but I wouldn't remember if I did – user574848 Nov 19 '18 at 6:15
• How many main 'routine' methods to solve number theory problems would you say there are? At higher levels do you think each question requires some different insight. – 3684 Nov 19 '18 at 6:19

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 $$\frac{n}{f(n)} = 1$$. 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

$$\boxed{\{199, 299, 399, 499, 599, 699, 799, 899, 999\}}$$

• Thank you for your answer, sorry I could only accept one answer. May I ask how you see the solution so quick? – 3684 Nov 19 '18 at 6:15