253 days until 2017

Consider the set of these 1008 consecutive odd numbers S = 3 , 5 , 7 , , 2017 S={3,5,7,\ldots,2017}

Now I randomly choose 673 distinct elements of S S . What is the probability (in percent) that there exists 3 co-prime numbers among the chosen ones?

Give your answer to the nearest integer.

Bonus : Generalize it!


The answer is 100.

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1 solution

Actually, I took inspiration from this problem.

Original problem: Randomly choose n + 1 n+1 numbers from the firtst 2 n 2n positive integers.Prove that there are 2 2 co-prime numbers in the choosen n + 1 n+1

Proof for the original problem: There must be at least 2 2 consecutive numbers in the choosen numbers. These 2 2 integers are co-prime (Q.E.D.)

Simple, right?

We can solve "253 days until 2017" similarly.

Generalized problem : Randomly choose 2 n + 1 2n+1 numbers from the consecutive 3 n 3n odd numbers. Prove that there are 3 consecutive odd numbers in the choosen 2 n + 1 2n+1 .

Proof : Suppose there are at most 2 consecutive, it is obvious that there are at most 2 n 2n choosen numbers(definitely false)

With

C C : choosen number

U U : unselected number

The sequences below which maximize the choosen 2 n + 1 2n+1 form the 3 n 3n odd numbers:

  • ( C U C ) (CUC) ( C U C ) (CUC) ( C U C ) (CUC) . . . ... ( C U C ) (CUC) (n groups)

  • ( C C U ) (CCU) ( C C U ) (CCU) ( C C U ) (CCU) . . . ... ( C C U ) (CCU) (n groups)

  • ( U C C ) (UCC) ( U C C ) (UCC) ( U C C ) (UCC) . . . ... ( U C C ) (UCC) (n groups)

Hence, there exists 3 consecutive odd numbers.(Q.E.D.). These 3 are also co-prime.

The answer is 100 %

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