The four numbers are (2*3*7)^98, (2*5*7)^70, (3*5*7)^42, and (2*7*11)^28.
Part (II) --
a) Any number which divides the 1st and 2nd must divide (2*7)^70.
b) Any number which divides the 1st and 3rd must divide (3*7)^42.
c) Any number which divides the 1st and 4th must divide (2*7)^28.
d) Any number which divides the 2nd and 3rd must divide (5*7)^42.
e) Any number which divides the 2nd and 4th must divide (2*7)^28.
f) Any number which divides the 3rd and 4th must divide 7^28.
We can ignore cases c, e and f, because they are all subsets of a.
So, we are looking for all numbers which divide (2*7)^70 or (3*7)^42 or (5*7)^42.
To compute these,
add
1) The number which divide 7^70 (ie, 71)
2) The number which divide (2*7)^70 but which are not a pure power of 7 = (70 powers of 2) * (71 powers of 7) = 4970
3) The number which divide
(3*7)^42 but which are not a pure power of 7 = (42 powers of 3) * (43 powers of 7) = 1806
4) The number which divide (5*7)^42 but which are not a pure power of 7 = (42 powers of 5) * (43 powers of 7) = 1806
Total number = 71 + 4970 + 1806 + 1806 = 8653
No time now to solve the 1st part. Perhaps somebody else will check my work, and finish the problem
2nd things 3rd (partial spoiler)
