TITLE: Making perfect cube from factors of $20!$
QUESTION [7 upvotes]: At least how many different factors of $20!$ must we choose so that we can always find some subset whose product is a perfect cube? For example, if we choose $\{2,3,5,7,11,13,17,19,22,26,34,38\}$, then no subset has a product that is a perfect cube, so the answer must be more than $12$.
We can write
$$20!=2^{18}\times 3^8\times 5^4\times 7^2\times 11\times 13\times 17\times 19$$
The total number of factors is $(18+1)(8+1)(4+1)(2+1)(1+1)^4=41040$, but I don't expect we need to use anything close to that.

REPLY [2 votes]: Seventeen.
There are sixteen factors that are primes or $2^3$ multiplied by a prime, and no non-empty product of these numbers is a perfect cube.
In the other direction, given a sequence of $17$ factors of $20!,$ we can map each factor $x_i=2^{n_{i,1}}3^{n_{i,2}}5^{n_{i,3}}\cdots 19^{n_{i,8}}$ to the tuple $n_i=(n_{i,1},\dots,n_{i,8})$ in the group $G=(\mathbb Z/3\mathbb Z)^8.$ By Olson's theorem (J.E. Olson, "A combinatorial problem on finite abelian groups I–II" J. Number Th. , 1 (1969) pp. 8–10; 195–199) $G$ has Davenport constant 17, which means some non-empty subsequence has total zero. This corresponds to a non-empty product of the factors being a perfect cube.