This problem is from IMC 2012, an international mathematics competition for undergraduate students. So, in IMC I have pretty bad time on problem 2, both day 1 and 2. After solving problem 2 on second day, I proceed to problem 3. There is only about 30 minutes left. The problem is like Pak Budi Surodjo’s style (Indonesia Number Theory coach in high school olympiad), that even Ronald Widjojo refused to attempt on this problem.

The idea is apparently very simple (compared to last year’s number theory problem). Well, you know that divides and is relatively prime to . This implies that must be divisible by . Apparently, there must be such that for all which means that such must be finitely many. We are done! (The progress you would make would probably not smooth as the flow in this paragraph, of course).

So, that’s it, a weird problem.

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