MathLabs

Problem 3

Let S={1,2,3,…,280}S=\{1,2,3,\ldots,280\}. Find the smallest integer n such that each n-element subset of S contains five numbers which are pairwise relatively prime.
Step 5 of 5: Force five members in one block
∣T∣=217 ⟹ ∣T∩(P∪A1∪A2∪A3∪B1∪B2)∣≥25>6⋅4|T|=217\ \Longrightarrow\ |T\cap(P\cup A_1\cup A_2\cup A_3\cup B_1\cup B_2)|\ge25>6\cdot4
Detailed analysis

The complement of the 88 selected numbers has 192 elements. Therefore a 217-element subset T meets their union in at least 25 elements. If every one of the six blocks contained at most four elements of T, the intersection would be at most 24. Hence one block contains five pairwise relatively prime elements. Combined with the obstruction, the minimum is n=217.