Without loss of generality, assume 1<a<b<c<d.

If a=3, then the maximum value for the LHS will be 1/4 + 1/5 + 1/6, however 1/4 + 1/5 + 1/6 < 2/3, thus a can only be 2.

By similar reasoning, we find that b=3,4,5.

For b=3,

there are 4 solutions: (2,3,7,42), (2,3,8,24), (2,3,9,18), (2,3,10,15)

For b=4,

there are 2 solutions: (2,4,5,20), (2,4,6,12)

For b=5,

there are no solutions.

Total: 6 solution sets

Q2) 120/ a^2 - 9 = 120/(a-3)(a+3)

Thus, a-3 and a+3 must both be factors of 120. Both factors must be positive. These 2 factors have a difference of 6. And, substitution of a, must yield a positive result from 120/(a+3)(a-3).

Hence we derive the following results. Do note that a = 3 is not possible as it will result in 120/0 - which is undefined.

2,8, a = 5

3,9, a = 6

4,10, a = 7

6,12, a = 9

24, 30, a = 27

Hence, a has 5 solutions.