MathLabs

Problem 1

Find the smallest positive integer n such that no arithmetic progression of 1999 real terms contains exactly n integers.
Step 3 of 5: Translate omission into a floor inequality
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k=⌊1999n+1⌋,k(n−1)+n≥2000k=\left\lfloor\frac{1999}{n+1}\right\rfloor,\qquad k(n-1)+n\ge2000
Detailed analysis

For a fixed n, choose k=floor(1999/(n+1)); this is the largest k for which the lower endpoint kn-k+1 is at most 1999. The next possible interval then starts beyond 1999 exactly when the upper endpoint (k+1)n-(k+1)+1 is at least 2000, equivalently k(n-1)+n is at least 2000.