I have been asked to answer the following question;
Let A = {2,3, ..., 50}, that is A is the set of positive integers greater than 1 and less than 51. Determine the smallest number $x$ such that every subset of A having $x$ elements contains at least two integers that have a common divisor greater than 1, and justify your answer.
My attempt at the solution is by defining a set;
$$S=[s\in A |s\in primes]$$
This defines the set containing 15 integers with no 2 primes containing a common divisor greater than one divisible by 1 and themselves. Thus, $x>15.$
Now I thought of taking;
$$P=[p|p\in primes]$$
And then continue by partitioning this set into 15 subsets. But I'm not too sure where to go from here.
Any help would be greatly appreciated.