14. How many sets of integers $(a, b, c)$ satisfy $2 \leq a \leq b \leq c$ and $a b c=2013 \times 2014$ ?
(2 marks)有多少組整數 $(a, b, c)$ 滿足 $2 \leq a \leq b \leq c$ 及 $a b c=2013 \times 2014$ ?
(2 分)
14. 90
14. The prime factorisation of $2013 \times 2014$ is $2 \times 3 \times 11 \times 19 \times 53 \times 61$. The problem is the same as counting the number of ways of distributing the 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 $c=1007$. We have a few cases to consider:
- The number of primes in the 3 groups are $4,1,1$ respectively (we shall abbreviated this as case $(4,1,1))$ - there are $C_{2}^{6}=15$ choices for the two singletons.
- Case $(3,2,1)$ - there are 6 choices for the singleton and then $C_{2}^{5}=10$ 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 $15+60+15=90$.