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Neporo4naja [7]
3 years ago
12

We select n + 1 different integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte

gers whose largestcommon divisor is 1.
Mathematics
1 answer:
just olya [345]3 years ago
6 0

Answer:

See answer below

Step-by-step 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

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