Consider the set of these 1008 consecutive odd numbers
Now I randomly choose 673 distinct elements of . 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!
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Actually, I took inspiration from this problem.
Original problem: Randomly choose n + 1 numbers from the firtst 2 n positive integers.Prove that there are 2 co-prime numbers in the choosen n + 1
Proof for the original problem: There must be at least 2 consecutive numbers in the choosen numbers. These 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 numbers from the consecutive 3 n odd numbers. Prove that there are 3 consecutive odd numbers in the choosen 2 n + 1 .
Proof : Suppose there are at most 2 consecutive, it is obvious that there are at most 2 n choosen numbers(definitely false)
With
C : choosen number
U : unselected number
The sequences below which maximize the choosen 2 n + 1 form the 3 n odd numbers:
( C U C ) ( C U C ) ( C U C ) . . . ( C U C ) (n groups)
( C C U ) ( C C U ) ( C C U ) . . . ( C C U ) (n groups)
( U C C ) ( U C C ) ( U C C ) . . . ( U C C ) (n groups)
Hence, there exists 3 consecutive odd numbers.(Q.E.D.). These 3 are also co-prime.
The answer is 100 %