1. What is the maximum number of numbers that can be chosen from the set $\{1,2, \ldots, 12\}$ so that the product of no three chosen numbers is a perfect cube?
Solution. 9: all except $4,9,12$. Note that to remove the cubes, we need to remove at least one element from each of the sets $\{1,2,4\},\{3,6,12\},\{2,4,8\},\{1,3,9\},\{2,9,12\}$, $\{3,8,9\},\{4,6,9\}$. Note that all numbers, except 9, are in no more than three of these seven triples. Therefore, if we remove two numbers different from 9, at least one triple will remain. If, however, we remove the number 9, then among the remaining triples there will be two non-intersecting $(\{1,2,4\},\{3,6,12\})$, so no matter which second number we remove, one of the triples will remain.