I took the term "factor" to mean "divisor with no remainder" as opposed to "prime factor").
With 60=2^2 x 3 x 5, a possible factor a student "i" might have (all else aside) is:
(2^j_i) x (3^k_i) x (5^l_i),
where, for n students, i goes from 1 to n. Each j_i, k_i, and l_i value can range from 0-120, 0-60, 0-60, respectively.
For a pair of students, i1, i2, to neither have the GCD of the pair, they must have 2 of the i,j,k exponents such that one student has a greater exponent and also a lesser exponent than the other. This is true because their GCD is the number formed using the "floor" of their exponents, when considered pairwise). E.g.: j_i1 > j_i2 and k_i1<k_i2. For a pair of students, this could happen in any of six ways, since each has their own set of three exponents.
So, the problem becomes one of finding the number of pairs of triplets, denoted {}, with this property, from set:
[{(0,0,0), (0,0,0)}, {(0,0,0), (0,0,1)}, ..., {(0,0,0), (120,60,60), {(0,0,1),(0,0,0)}, {(0,0,1),(0,0,1)}, ..., {(120,60,60) , (120,60,60)}]
and dividing the count by 2, since each satisfactory pair will be counted twice.
Being lazy, I wrote a program to count these:
maybe a soln.....
