Problem 1
Find the smallest positive integer n such that no arithmetic progression of 1999 real terms contains exactly n integers.
Step 4 of 5: Bound the first possible answer
Detailed analysis
If the inequality fails, then 1999(n-1)/(n+1)+n<2000. Simplifying gives n squared less than 3999, hence n is at most 63. Therefore it is enough to check n from 64 onward until the inequality first holds.