A divisive year

2016 can be factorized as $2^{5}\ast 3^{2}\ast7^{1}$ . If we need to arrange this into three factors, $(a{_1},a{_2},a{_3})$ then we have $a{_1}*a{_2}*a{_3}$ = 2016. So each of $a{_1}, a{_2}, a{_3}$ can be expressed as powers of prime factors, $a{_i} = 2^{x{_i}} \ast 3^{y{_i}} \ast 7^{z{_i}}$ . We have three equations $\sum_{i=1}^3 x{_i} = 5, \sum_{i=1}^3 y{_i} = 2 , \sum_{i=1}^3 z{_i} = 1$ . Its easy to see the total number of possibilities are $7c{_2} \ast 4c{_2} \ast 3c{_2}$ = 378. Since the problem states 'integers', we need to consider the possibilities where 2 among $a{_i}$ 's are negative. So that gives us $3c{_2}$ more possibilities. So total = $(3c{_2} + 1) \ast 378$ = $\boxed{1512}$