MathLabs

Problem 1

Find the smallest positive integer n such that no arithmetic progression of 1999 real terms contains exactly n integers.
Step 5 of 5: Check the boundary
Previous stepNext step
30⋅64+64,…,28⋅69+69<2000,28⋅69+70=2002≥200030\cdot64+64,\ldots,28\cdot69+69<2000,\qquad 28\cdot69+70=2002\ge2000
Detailed analysis

For n=64,65,66,67,68,69 the values of k(n-1)+n are respectively 1954, 1985, 1951, 1981, 1944, 1973, all below 2000. For n=70, k=floor(1999/71)=28 and the value is 28*69+70=2002. Thus 70 is the smallest omitted count.