There are $n$ students with a different student number. Each student number is a positive factor of $60^{60}$ In a school there are $n$ students, each with a different student number. Each student number is a positive factor of $60^{60},$ and the GCD of two student numbers is not a student number in the school. Find the greatest possible value of $n$. 
I am completely lost in this problem. What should I do to solve it?
So far what I have done: $60^{60} = 2^{120}*3^{60}*5^{60}$
Each student number is of form: $2^{a}*3^{b}*5^{c} (0 \leq a \leq 120, 0 \leq b \leq 60,0 \leq c \leq 60)$
 A: There's a slight/arguable ambiguity in the condition as to whether the 3 students need to be distinct, or the 3rd is allowed to be either the first or the second. We take the latter's interpretation.   

Consider the lattice of points $(a, b, c)$ where $0 \leq a \leq 120, 0 \leq b \leq 60,0 \leq c \leq 60)$.
We define a partially ordered set where $(a_1, b_1, c_1) \leq (a_2, b_2, c_2 ) $ iff $ a_1 \leq a_2, b_1 \leq b_2 , c _1 \leq c_2 $.
The question asks for the size of the largest antichain.   
Hint: By Dilworth's theorem states that the size of the largest antichain is equal to the minimum number of chains into which the set can be partitioned.

 Observe that $X = \{ (120-b-c, b, c ) \}$ is an antichain of 3600 elements.
 Conversely, we have 3600 chains that cover the lattice, namely $(0,b,c) -> (120, b, c)$.

Hence, the largest antichain has size 3600. 
A: HINT
We can start with two numbers, e.g. $5$ and $3.$ To be sure that GCD is not a student's number, combine increasing powers of one with decreasing powers of the other
$$5^{60}3^{0}, 5^{59}3^1,\cdots,5^{0}3^{60}.$$ We have  $61$ student numbers $$5^{60-k}3^k$$ until now.
