50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Answer: a) any pair of divisors of the number $2010$ with product $2010$, for example $a_1=2$, $a_2=1005$; b) $2$, $3$, $5$, $6$, $10$, $15$, $134$, $201$, $339$, $402$, $670$, $1005$.

The main idea is: if $n$ is not a square of an integer and $d$ is its divisor, $1<d<n$, then $d^{\prime }=\frac{n}{d}$ is also a divisor of $n$, for which $1<d^{\prime }<n$ and $d\ne d^{\prime }$.

a) In such a way, for $a_1$ we can take any divisor of $2010$, that differs from $1$ and $2010$, and then put $a_2=\frac{2010}{a_1}$.

b) Analogously, we can write out pairwise distinct pairs of divisors of $2010$, that satisfy the condition above. The answer was given before.