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 $n=f(n)$?
Clash Royale CLAN TAG #URR8PPP up vote 3 down vote favorite 1 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? elementary-number-theory contest-math share | cite | improve this question edited Nov 19 at 12:31 amWhy 191k 27 223 437 asked Nov 19 at 5:36 3684 30 6 New contributor 3684 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. in the definition of