Problem:
Find the number of ordered pairs of integers such that the sequence
is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Solution:
Answer (228):
Neither nor can be 6 or 20 because either of those numbers would make an arithmetic progression with either the least three numbers or the greatest three numbers. Therefore and must be chosen from the remaining 22 values between 7 and 29 not including 20. Setting results in the only arithmetic progression that contains two of the three least values. Because 20 is already excluded, no arithmetic progression can be formed using exactly two of the three greatest numbers. All other possible arithmetic progressions must start with one of 3,4 , or 5 and end with one of 30,40 , or 50 . The difference of the starting and ending terms must be divisible by 3 , which yields three potential pairs for starting and ending numbers of the arithmetic progression of length four. Using 3 and 30 , only gives an arithmetic progression. Using 4 and 40 , only gives an arithmetic progression. The arithmetic sequence starting with 5 and ending with 50 would contain 20 and has already been excluded. Therefore the requested number of ordered pairs is .
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions