Jutaro wrote: The Central American Olympiad is an annual competition. The ninth Olympiad is held in 2007. Find all the positive integers $n$ such that $n$ divides the number of the year in which the $n$-th Olympiad takes place. If the 9th Olympiad was held in 2007 it means the 1st Olympiad was held in 1999. And in general the $n$-th Olympiad is held in $1998+n$. We ask for what $n$ does $n | (1998+n)$ which is true if and only if $n|1998$. So all positiver integer $1998+n$ such that $n|1998$ are all the year the Olympiads that you seek.

Since the 9th one was held on 2007, in general, the nth olympiad is held in the year 1998 + n. Therefore we need to find all n that satisfy 1998 + n = 0 (mod n ). Therefore 1998 = 0 (mod n). Therefore all n we find are such that n/1998. In other words, the divisors of 1998.

Using the theory of prime factors, I can highlight that 1998 acts as 0, which could say that if 1998 has a prime factor then that prime factor will appear again by adding that amount of the prime factor. For example 2 divides 1998 so 2 divides 1998+2 3 divides 1998 so 3 divides 1998+3 and so progressively. So we have that 1998 has 16 divisors so if we add those 16 to 1998 we will then have 16 answers. Answers: 1|1999 2|2000 3|2001 6|2004 9|2007 18|2016 27|2025 37|2035 54|2052 74|2072 111|2109 222|2220 333|2331 666|2664 999|2997 1998|3996