Answer:
See answer below
Explanation:
From the set
{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.
Then
it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers
if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3 → the largest common divisor of that number with 3 is 1
If the set includes any other prime number → the largest common divisor of that with any other is 1
For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :
N = N₂*n₂ , since n₂ ≥ 2 → N₂ < N
then the even N₂ also should be contained in the set
therefore also for N₂
N = N₃*n₃ → N₃ < N₂
therefore if we continue , we would obtain a number even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1