A set of integers whose elements all divide $2015^{200}$ but do not divide each other

Note that any two distinct elements $5^{p_1} 13^{q_1} 31^{r_1}$ and $5^{p_2} 13^{q_2} 31^{r_2}$ are incomparable (neither divides the other) if $p_1 + q_1 + r_1 = p_2 + q_2 + r_2$. So a mutually incomparable set is given by $$ S_n = \left\{5^p 13^q 31^r \;\big\vert\; p+q+r=n;\; 0\le p,q,r \le 200\right\} $$ for any $n$. The size of this set is maximal when $n=300$ (*). To show that this is the largest mutually incomparable set, we can apply Dilworth's theorem, which says that if an antichain $A$ has cardinality equal to that of a partition $P$ of the set into chains, then that antichain is maximal. A corresponding partition here is given by (the transitive closure of) the following equivalence relation: if $x=5^p 13^q 31^r$, then

• If $p+q+r\equiv 0$ (mod $3$), $x \sim 5x$,
• If $p+q+r\equiv 1$ (mod $3$), $x \sim 13x$, and
• If $p+q+r\equiv 2$ (mod $3$), $x \sim 31x$.

The equivalence relation links every triple $(p,q,r)$ to a unique triple with $p+q+r=300$, and does so while staying inside the box $0\le p,q,r \le 200$. (The latter property is key, since it distinguishes, say, $S_{300}$ from the much smaller $S_{200}$.)

Note on evaluating $|S_n|$:

Only the restrictions $p,q,r\ge 0$ apply for $0 \le n\le 200$, so the number of ways to choose $(p,q,r)$ summing to $n$ in that range is $$|S_n|=1+2+\ldots+n+(n+1)=\frac{1}{2}(n+1)(n+2).$$ For $200<n \le 300$, we have to subtract out cases where one of the terms (and it can be only one) is larger than $200$: there are $$1+2+\ldots+(n-200)=\frac{1}{2}(n-200)(n-199)$$ cases where each of $p,q,r$ is greater than $200$. So the total size is $$ |S_n|=\frac{1}{2}(n+1)(n+2)-\frac{3}{2}(n-200)(n-199) $$ in this range. (So $|S_{300}|=30301=|S_{299}|+1$.) Finally, for $n>300$, we can use symmetry, because $|S_{n}|=|S_{600-n}|$.