Hong Kong Preliminary Selection Contest

The prime factorisation of $2013\times 2014$ is $2\times 3\times 11\times 19\times 53\times 61$. The problem amounts to counting the number of ways of distributing these 6 primes into 3 groups such that each group contains at least one prime. For example, if the groups are $\{2,3,11\}$, $\{19,53\}$ and $\{61\}$, then the corresponding factors are 66, 1007 and 61. By ordering the factors, we get $a=61,b=66$ and $c=1007$, and hence $(a,b,c)=(61,66,1007)$ is a possible triple. We have a few cases to consider:

The number of primes in the 3 groups are 4, 1, 1 respectively (we shall abbreviate this as case (4, 1, 1)) — there are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ choices for the two singletons.

Case (3, 2, 1) — there are 6 choices for the singleton and then $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ choices for the doubleton, leading to $6\times 10=60$ ways in this case.

* Case (2, 2, 2) — there are 5 ways to choose a partner for the prime 2, and 3 ways to choose a partner for one of the remaining primes. Afterwards, the two primes left form the last group automatically. Hence there are $3\times 5=15$ ways to do so.

It follows that the answer is $15+60+15=90$.