By constructing the set in this way, you can guarantee that no three integers are consecutive terms of an arithmetic progression.
Yes, it is possible to choose 1983 distinct positive integers, all less than or equal to 100,000, such that no three of them form consecutive terms of an arithmetic progression. This is a known result in number theory and combinatorics.
To construct such a set, you can use a combinatorial argument. Consider the set of positive integers from 1 to 1983. Now, for each integer in this set, assign a unique prime number to it. The nth integer in the set will be assigned the nth prime number. This ensures that every integer has a unique prime factorization.
Now, consider the set of prime numbers less than or equal to 100,000. There are more than enough prime numbers in this range to assign to the first 1983 positive integers. Since every integer in the original set has a unique prime factorization, no three integers can form an arithmetic progression because they would have to share a common prime factor, which is not possible.
Therefore, by constructing the set in this way, you can guarantee that no three integers are consecutive terms of an arithmetic progression.
Question
Is it possible to choose 1983 distinct positive integers, all less than or equal to 100,000, no three of which are consecutive terms of an arithmetic progression?